Chem Module 5

Rates can be measured either as an

average rate during a longer Δt time interval or as an instantaneous rate during a very short Δt time interval.

Explain why reaction rate is affected by a temperature change.

b)As the temperature is changed, the kinetic energy of the reacting particles is changed and this changes the probability of particles colliding with sufficient energy to undergo the reaction.

If the Kc is ≈ 1, the equilibrium mixture will contain a significant concentration of

both products and reactants.

AVERAGE rate equals

change in {} / Change in time

Effect of pressure change: (2)

change the volume of a reaction which involves a change in total moles of gas as the reaction takes place. This volume change actually changes the partial pressure (P = nRT/V) of the gases present which effectively changes the concentrations of those gases. This effect can be seen in the catalytic methanation reaction which most effectively takes place at moderately high temperature and moderately high pressure.

These reactions can be "pushed" in the forward or reverse direction by

changing the conditions so as to preferentially form a particular product or reactant. An example of such a reversible equilibrium reaction is steam methane reforming one which is widely used in the chemical industry and is shown below. CH4 (g) --> 3 H2 (g) <---> Ch4 (g) + H20 (g) Can be written the other way too

The frequency of particle collisions depends on both their...

concentration and temperature of the system

graphs of reaction rate

concentration of product formed (below, top) is plotted versus time

Normally, an increase in the concentration of a reactant causes a

corresponding increase in the rate of reaction since this provides more of those reacting particles to undergo the reaction, but occasionally the reaction rate is unaffected by a concentration increase in certain reactants.

Graph of reaction rate

decreasing concentration of reactant versus time.

Use the [O2] data from the table to calculate the instantaneous rate early in the reaction (0 secs to 300 sec).

find early instantaeous; will be the largest rate value Answer: rate = Δ[O2] / Δt = (0.010 - 0) / 300 - 0 = 3.33 x 10-5 mol/L•s

Higher concentration of particles assures a greater

frequency of this greater number of particles colliding

Using the equilibrium constant

-To predict whether an equilibrium favors formation of products or reactants. -To predict whether a reaction not at equilibrium will proceed toward products or reactants. -To predict equilibrium concentrations for any set of starting concentrations.

A sample of wood from an ancient tomb was found to contain 15.7 % 14C content as compared to a present-day sample. The t1/2 for 14C is 5720 yrs. What is the age of the wood?

0.693 = k t1/2 0.693 = k (5720 yrs) k = 0.693 / 5720 = 1.2115 x 10-4 ln [A] - ln [A]0 = - k t ln 15.7 - ln 100 = - (1.2115 x 10-4) t 2.7537 - 4.6052 = - (1.2115 x 10-4) t t (age) = - 1.8515 / - 1.2115 x 10-4 = 15,283 yrs

A sample of paper from an ancient scroll was found to contain 39.5 % 14C content as compared to a present-day sample. The t1/2 for 14C is 5720 yrs. What is the age of the scroll?

0.693 = k t1/2 0.693 = k (5720 yrs) k = 0.693 / 5720 = 1.2115 x 10-4 ln [A] - ln [A]0 = - k t ln 39.5 - ln 100 = - (1.2115 x 10-4) t 3.6763 - 4.6052 = - (1.2115 x 10-4) t t = - 0.9289 / - 1.2115 x 10-4 = 7667.4 yrs

Shifting Equilibrium reactions:: Since an equilibrium reaction does not go to completion, significant reactant is left unreacted and essentially wasted. It would be helpful if an equilibrium reaction could be pushed toward the products side so that the yield of products could be maximized. This can be done since there are three methods of shifting equilibrium reactions:

1. Changing the concentrations by adding or removing reactants or products. 2. Changing the partial pressure of gaseous reactants or products by changing the volume. 3. Changing the temperature.

The instantaneous rate near the beginning of the reaction will be ___________ than the instantaneous rate at the end of the reaction or the average rate

greater

A higher temp assures that these faster movving particles will collide with

greater frequency However, only collisions with proper alignment of the colliding particles in which the particles have greater than the activation energy will be a productive one resulting in product being formed.

Effect of concentration changes (1)- remove or add reactant One way to increase the yield of product in an equilibrium reaction is to change the concentration of the equilibrium mixture by removing a product or adding a reactant. This concentration change causes the equilibrium to shift briefly to partially restore the equilibrium concentration of the removed or added substance. Suppose, for instance that the water vapor (H2O(g)) were removed at equilibrium from the catalytic methanation reaction below: CO (g) + 3 H2 (g) <==> CH4 (g) + H2O (g)

his would cause the reaction to shift briefly so as to partially restore the water vapor concentration which means that for a brief period of time the forward reaction rate exceeds the reverse reaction rate converting CO and H2 to CH4 and H2O until equilibrium is again established. Incidentally, this shift has produced more CH4, increasing the yield of this desired product. What this change has accomplished is the production of a greater yield of a desired product by the removal of a common and ordinary product (H2O). So in summary, removal of a product from a reaction at equilibrium, caused the reaction to briefly shift in the forward direction (by increased rate of the forward reaction) to produce more of the removed product to partially restore the concentration of that material.

Reaction rate is the

increase in molar concentration of product per unit of time or the decrease in molar concentration of reactant per unit of time usually expressed in moles per liter per second (mol/L•s)

Increasing the surface area of the particles of solid reactants by grinding the solid into a powder causes an

increase in the rate of reaction since solids react only at their surfaces

Rate law

is a mathematical equation that describes the dependence of the reaction rate on the concentrations of some of the reactants. The rate law will have the following form for the hypothetical reaction where k is the rate constant (which is constant only for a given temperature and solvent) and [ ] indicates molar concentration. aA + bB → cC + dD rate = k [A]x [B]y

If the reaction has a positive Heat of Reaction (+∆H)

it will be Endothermic

If the reaction has a negative Heat of Reaction (-ΔH),

it will be Exothermic.

rate constant related to the temperature is

k = A e-Ea/RT

Ex of half life

ln 50 - ln 100 = - k t1/2 3.912 - 4.605 = - k t1/2 0.693 = k t1/2

Half-life values are characteristic of the reactions and reactants and are especially useful for describing the decay of radioactive nuclei. The two equations shown in bold above (and below) can be used to solve most first-order rate problems, using the right equation only for half-life times where 50% of the reactant has reacted and the left equation for any time or % of reaction:

ln [A] - ln [A]0 = - k t 0.693 = k t1/2

EX: Determine the half-life of a radioactive nucleus if 90% of the material has decayed in 366 mins (remember that this means only 10% is remaining).

ln [A] - ln [A]0 = - k t ln 10 - ln 100 = - k (366 mins) 2.3026 - 4.6052 = - k (366) - k (366) = - 2.3026 k = 2.3026 / 366 = 6.29 x 10-3 0.693 = k t1/2 0.693 = (6.29 x 10-3) t1/2 t1/2 = 0.693 / 6.29 x 10-3 = 110.17 mins

Determine the decay constant (k) and the half-life of a radioactive nucleus if 75% of the material has decayed in 400 days.

ln [A] - ln [A]0 = - k t Remaning - total = ln 25 - ln 100 = - k (400 days) 3.2189 - 4.6052 = - k (400) - 1.3863 = - k (400) k = - 1.3863 / - 400 = 3.47 x 10-3 0.693 = k t1/2 0.693 = (3.47 x 10-3) t1/2 t1/2 = 0.693 / 3.47 x 10-3 = 199.7 days

A reaction with relatively large Eact will be a

nonspontaneous reaction requiring the input of significant energy to cause the reaction to occur.

Writing Kc for a heterogeneous equilibrium

omit the concentration term for solids (s) or liquids (liq) or solvents (since the concentrations of these materials are constants), but NOT for gases (g) or solutions (aq). So the Kc expressions for the above equilibria would be as shown below: 4 H2 (g) + Fe3O4 (s) <==> 3 Fe g(s) + 4 H20 (g) Kc = [H2O]4 ---------------- [H2] 4 H20 (liq) + 2 Cl2 (g) <==> 4 HCl (aq) + O2(g) Kc= [HCl]4[O2] ------------------ [Cl2]2

If the Kc is large then the equilibrium mixture must be predominantly...

products 2.0 M

Enzymes are

protein catalysts present in living systems that promote reactions, often in a fraction of a second and at mild conditions of temperature and pressure, that would be impossible otherwise. An automobile catalytic converter makes use of platinum, palladium and rhodium metal catalysts to convert some of the noxious pollutants present in automobile exhaust to less harmful substances.

Use the [CH3CHO] data from the table to calculate the instantaneous rate late in the reaction (15000 secs to 20000 secs).

rate = - Δ[ CH3CHO] / Δt = (0.0043 - 0.0056) / 20,000 - 15,000 = 2.60 x 10-7 mol/L•s

Use the [CH3CHO] data from the table to calculate the average rate over the measured time interval from 0 to 20000 secs.

rate = - Δ[ CH3CHO] / Δt = - (0.0043 - 0.0500) / 20,000 - 0 = 2.285 x 10-6 mol/L•s

Use the [CH3CHO] data from the table to calculate the instantaneous rate early in the reaction (0 secs to 1200 sec).

rate = - Δ[ CH3CHO] / Δt = - (0.0300 - 0.0500) / 1200 - 0 = 1.67 x 10-5 mol/L•s

If the Kc is very small then the equilibrium mixture must be predominantly...

reactants ex 6.8 x 10 . -16

the instantaneous rate decreases (reaction is slower) as the reaction proceeds

since the concentration of reactants becomes smaller due to consumption of reactants as the reaction takes place and the rate finally becomes 0 as the reaction stops.

A reaction with relatively small Eact will be a

spontaneous reaction requiring the input of very little energy (enough that it is supplied by that available at room temperature) to cause the reaction to occur.

how the value of Kc can be determined if the equilibrium concentrations are given.

started with 1.000 mole of CO and 3.000 mole of H2 and we knew only the equilibrium amount of H2O (0.387 mole), we could determine the equilibrium amounts of the other substance since according to the reaction stoichiometry, for every mole of H2O formed, a mole of CH4 must be formed and a mole of CO must have reacted and 3 moles of H2 must have reacted. So the equilibrium amounts of the other substances must be: 0.387 mole of H2O formed = 0.387 mole of CH4 formed 0.387 mole of H2O formed = 0.387 mole of CO reacted so CO = 1.000 - 0.387 = 0.613 mole of CO 0.387 mole of H2O formed = 3 x 0.387 mole of H2 reacted so H2 = 3.000-3(0.387) = 1.839 mole of H2

This indicates that for any chemical equilibrium reaction the equilibrium constant, Kc, is equal to

the product of the equilibrium concentrations (M = mol/L) of the products divided by product of the equilibrium concentrations (M = mol/L) of the reactants with each concentration raised to a power equal to their reaction coefficient. This expression, Kc, is constant for a given temperature.

The third requirement for a reaction to occur is that

the reactant particles must be raised to a higher potential energy level (activation energy) before they can be converted to products which lie at a lower potential chemical energy level than the initial reactant potential energy level.

Kinetics

the study of rates of reactions and how they are influenced by certain factors like concentration, temperature, catalysts and surface area. In general, reaction rates increase with an increase in temperature since the higher temperature provides a higher kinetic energy for the reacting particles

overall reaction order

the sum of those exponents

Le Chatelier's Principle

when a system at equilibrium is disturbed by a change of temperature, pressure or concentration, the system shifts in equilibrium composition so as to counteract the change imposed on it.

ln [A] - ln [A]0 = - k t

where ln [A]0 is the natural logarithm of the initial concentration, ln [A]is the natural logarithm of concentration at time t and k is the rate constant.

2 N2O5 (g) → 4 NO2 (g) + O2 (g) rate = Δ[O2] / Δt or rate = - Δ[N2O5] / Δt

where Δt indicates change in time (which is defined as tfinal - tinitial) and Δ[ ] indicates change in molar concentration (which is defined as concfinal - concinitial). The negative sign in the N2O5 rate expression is used anytime a reactant concentration is referred to, so as to make the rate positive since the Δ[N2O5] change is a negative value.

First order rate law is used to determine the half life of a reactant

which is defined as the time required for one-half of a reactant to undergo the reaction and takes the form of the following equation which is the first order rate law for A = 50% A0:

Effect of Catalyst

A catalyst increases the rates of both the forward and reverse reactions in a chemical equilibrium system and, thereby, more quickly brings the system to equilibrium. The catalyst does NOT, however, have any effect on the composition of the equilibrium mixture or the value of the equilibrium constant, Kc. This may not seem like much of an advantage but consider the case of the following reaction which has a Kc value of 1.7 x 1026: This large Kc value indicates that an equilibrium mixture of this system should contain mostly SO3 but when sulfur is burned in excess air, very little of the initially-formed SO2 is converted to SO3 by this reaction since the rate of the forward reaction is very slow. However, when a Pt or V2O5 catalyst is added, the reaction efficiently forms an equilibrium mixture composed of mostly SO3, in agreement with the expected amount predicted from Kc. Here, as in all cases, the composition of the equilibrium mixture is unaffected by the catalyst, but the rate of attainment of equilibrium is greatly increased.

Effect of Catalyst

A catalyst is any substance that increases the rate of a chemical reaction without itself being consumed. This occurs by the catalyst combining with the reactants to form an intermediate that is able to react more readily to form the product. During the reaction to form the product, the catalyst is regenerated. Catalysts are so very useful since they are ideally reusable, are only necessary in very small amounts and are very specific in the reaction they catalyze and the products they form.

Second-Order Reaction

A few reactions are second order with respect to some reactants. For a second-order reaction, the rate of reaction is proportional to the square of the concentration of one of the reactants. Rate Law: r = k [A]2 the rate constant, k, has units of L/mol•s.

PLUG IT IN PLUG IT IN

A1, B1, A2, B2 all found in the chart given Rate 1 Rate 2, given in the chart Cancel out all common terms

Effect of Temperature

Almost all reaction rates are exponentially increased as the temperature is increased since a higher temperature imparts greater kinetic energy to the reactants which allows the particles to come into contact with one another more frequently and with greater probability of having sufficient energy to undergo the reaction

The Kc value can also be used to predict whether a reaction not at equilibrium will proceed toward products or reactants as it comes to equilibrium.

As an example let's look at placing a catalytic methanation mixture with the following composition placed in a container at 927°C with a catalyst: 0.0400 M CO, 0.0400 M H2, 0.00200 M CH4 and 0.00200 M H2O. Will this reaction, shown below with its equilibrium constant, proceed to form more CO and H2 (toward the reactants) or more CH4 and H2O (toward the products) as it comes to equilibrium?

Average rate vs instantaneous rate

Average rate is usually determined by using initial and final concentrations and initial and final times so that the result measures the rate over the entire reaction (the average rate). Instantaneous rate is determined by using concentrations and times near the beginning or end (or any other time interval) of the reaction so that the result is the rate during a short time span (the instantaneous rate).

rate laws must be determined experimentally and it should be understood that the exponents x and y are not necessarily equal to the coefficients (a and b) of reactants A and B in the balanced reaction equation. For example for the reaction:

Br- + BrO3- + 6 H+ → 3 Br2 + 3 H2O the rate law is: rate = k [Br-] [BrO3-] [H+]2

When the reaction is written in the reverse order (which is the reaction for steam methane reforming), the Kc expression is inverted as shown below:

CH4 (g) + H20 (g) <===> CO (g) + 3H2 (g) [CO] [H2]3 Kc= ________________ [CH4] [H20]

Use the [O2] data from the table to calculate the average rate over the measured time interval from 0 to 3000 secs.

Change in oxygen conc over the change in time O2 #2 - O2 #1 / t2 -t1 =m/L/s Value will be in between the early and late Answer: rate = Δ[O2] / Δt = (0.054 - 0) / 3000 - 0 = 1.80 x 10-5 mol/L•s

Kc Equilibrium constant Problem Ex: Catalytic Methadation Reaction If you know the -reaction -Coefficients -beginning conc of reactants -atleast one final conc of product then you can find Kc

Co + 3H2 <==> CH4 + H20 Kc= [CH4] [H20] _________________________ [CO] [H2]3 Start with 1.00 mol of CO & 3.00 mol of H2 0.387 mol of CH4 & 0.387 mol of H2O 1.00- 0.387 mol of CO 3.00 - 3(0.387) mol of H2 In Mol but ned to change to Liters so / by 10 Kc= (0.387/10) (0.387/10) ___________________________________ (0.613/10) (1.893/10)>3 Kc= 3.93

homogeneous equilibria

Each reactnant and product is in the same physical state: The two equilibria which we have discussed steam methane reforming and catalytic methanation (shown below) are examples of homogeneous equilibria where all the reactants and products are present in a single phase; in these two cases all materials are in the gas phase. Ch4 (g)+ H20 (g) <==> CO (g) + 3H2 (g) CO (g) + 3 H2 (g) <==> CH4 (g) + H2O (g)

How instantaneous rates and average rates compare

Early instantaeous rate has the highest conc of reactants > average rate > late inst has lowest conc of react

First-Order Rate Law

First-order reactions are the most common type with the rate law with respect to time especially describing the behavior of radioactive materials, one of the most common first order reactions. For a first-order reaction, the rate law with respect to time is: ln [A] - ln [A]0 = - k t

In order to study reaction rates, the concentrations of reactants or products must be measured as the reaction takes place. For gas reactions, most of which involved a change in the number of moles present, by monitoring the change in pressure, at constant volume and temperature, one can obtain the reaction rate.

Following the change in absorption of light by one of the reactants or products is another common method of measuring reaction rates.

How the reversible reaction works

For the catalytic methanation reaction, if we place 1.00 mole of CO and 3.00 mole of H2 in a container under proper reaction conditions, the rate of the forward reaction of CO and H2 is initially large due to the relatively large concentrations of these substances but the forward rate steadily decreases as these substances are consumed. The rate of the reverse reaction of CH4 and H2O is initially small due to the relatively low concentrations of these substances but the reverse rate steadily increases as these substances are produced. Eventually, the forward reaction rate decreases and the reverse reaction rate increases to the point where they are equal and the system is at equilibrium. From this point on, no net concentration changes are observed since the forward and reverse processes continue at equal rates. Since the forward and reverse reactions continue (at equal rates) this is called a dynamic (active) equilibrium even though there is no overall change in the concentrations which might lead one to believe that it is a static (unchanging) situation.

If the temperature is decreased on a reaction at equilibrium that has a +ΔH0, the reaction will briefly shift in the direction that produces some heat [+ΔH0 indicates an endothermic (heat absorbing) forward reaction] so this reaction will shift in the reverse direction. Simultaneously, this reverse shift will decrease the concentration of the products (in the numerator of Kc) and increase the concentration of the reactants (in the denominator of Kc) which will decrease the value of Kc.

Heat of Reaction Temp Change Equilibrium Shift Kc Change +ΔH0 Increase Forward Increase +ΔH0 Decrease Reverse Decrease -ΔH0 Increase Reverse Decrease -ΔH0 Decrease Forward Increase

Summary of Concentration changes:

If the concentration of a material in a reaction at equilibrium is increased, the reaction will briefly shift in the direction that consumes some of that added material to re-establish the equilibrium. If the concentration of a material in a reaction at equilibrium is decreased, the reaction will briefly shift in the direction that produces some of that removed material to re-establish the equilibrium.

Summary of pressure changes

If the pressure is increased on a reaction at equilibrium that involves a change in the number of moles of gas, the reaction will briefly shift in the direction that decreases the pressure by going toward the side with the lesser moles of gas. If the pressure is decreased on a reaction at equilibrium that involves a change in the number of moles of gas, the reaction will briefly shift in the direction that increases the pressure by going toward the side with the greater moles of gas. Remember that only gases are considered in determining the effect of pressure on an equilibrium reaction.

If the temperature is increased on a reaction at equilibrium that has a +ΔH0, the reaction will briefly shift in the direction that uses up some of the added heat [+ΔH0 indicates an endothermic (heat absorbing) forward reaction] so this reaction will shift in the forward direction. Simultaneously, this forward shift will increase the concentration of the products (in the numerator of Kc) and decrease the concentration of the reactants (in the denominator of Kc) which will increase the value of Kc.

If the temperature is decreased on a reaction at equilibrium that has a -ΔH0, the reaction will briefly shift in the direction that produces some heat [-ΔH0 indicates an exothermic (heat producing) forward reaction] so this reaction will shift in the forward direction. Simultaneously, this forward shift will increase the concentration of the products (in the numerator of Kc) and decrease the concentration of the reactants (in the denominator of Kc) which will increase the value of Kc.

The equilibrium reaction below has the Kc = 0.250 at 25°C. If the temperature of the system at equilibrium is decreased to 0°C, how and for what reason will the equilibrium shift. Also show and explain how and why the Kc value will change. 2 SO2 + O2 <==> 2 SO3 ChangeH = -94.1 kJ

If the temperature is decreased on this reaction at equilibrium which has a -ΔH0, the reaction will briefly shift in the direction that produces some heat [-ΔH0 indicates an exothermic (heat producing) forward reaction] so this reaction will shift in the forward direction. Simultaneously, this forward shift will increase the concentration of SO3 (in the numerator of Kc) and decrease the concentration of SO2 and O2 (in the denominator of Kc) which will increase the value of Kc.

Summary of Temp changes

If the temperature is increased on a reaction at equilibrium that has a -ΔH0, the reaction will briefly shift in the direction that uses up some of the added heat [-ΔH0 indicates an exothermic (heat producing) forward reaction] so this reaction will shift in the reverse direction. Simultaneously, this reverse shift will decrease the concentration of the products (in the numerator of Kc) and increase the concentration of the reactants (in the denominator of Kc) which will decrease the value of Kc.

Ex of transition state

NO (g) + O3 (g) → [ O-N-----O-----O-O ] → NO2 (g) + O2 (g)

Reversible reactions

Not all chemical reactions completely convert reactants to products. Many reactions are said to be reversible reactions since as soon as some product molecules are formed, these begin to recombine to form reactant molecules while additional reactant molecules continue to convert to products. The forward reaction (reactants → products) continues to occur as the reverse reaction (products → reactants) also occurs until the reaction becomes a mixture of products and reactants.

Explain why reaction rate is affected by a reactant concentration change.

(a) A change in reactant concentration changes the number of particles colliding which changes the probability of the particles undergoing the reaction.

(c) Explain why powdered zinc reacts faster than a chunk of zinc.

(c) Solid particles only react at their surface and since powdered zinc particles have a greater surface energy than a chunk of zinc the powdered zinc will react faster.

(d) Explain how a catalyst works.

(d) A catalyst speeds up the rate of a chemical reaction by causing formation of an alternate transition state with lower Energy of Activation, thereby allowing more reactant molecules to form the Transition State at lower energy.

(e) Explain why reactions in solution are usually faster than heterogeneous reactions.

(e) The reactants particles can collide with one another much more readily in solution whereas particles in heterogeneous reactions can only collide by way of their surfaces. An increase in collisions causes an increase in reaction rate.

(f) Explain why stirring speeds up the rate of heterogeneous reactions.

(f) Stirring heterogeneous reactions increases the collisions between particles which speeds up the rate of reaction.

What is drawn to show transition states and the reactions in which they are involved?

Potential energy diagrams- endothermic and exothermis

So in general if a reversible reaction is not at equilibrium and the values of Qc and Kc are known we can predict that the following will occur as the reaction proceeds to equilibrium:

Qc > Kc: the reaction will proceed to the left (in the direction of the reactants) Qc < Kc: the reaction will proceed to the right (in the direction of the products) Qc = Kc: the reaction is at equilibrium

-First: FIll in equation rate 1 = k[A1]x [B1]x ___________________________ rate 2 = k [A2]x B2]x If there are more than just A, B then do C, D etc as well -Cancel common terms 3. DO math 4. x or y = ? with respect to whatever element was used the whole time 5. Once finding value for x, y, z etc then plug into origianl rate :Rate = k [ClO-]x [I-]y [OH-]z 6. Determine the value of k

Rate = k [A]x [B]y 1. 1.8 x 10-4 = k[0.0030]x [0.0010]y [1.00]z ___________________________________ 3.6 x 10-4 = k [0.0030]x [0.0020]y [1.00] z 2. 1.8 x 10-4 = k [0.0010]x ___________ ____________ 3.6 x 10-4 = k[0.0020]x 3. 0.50 = (0.5)x 4. x =1 with respect to ClO- 5. k [ClO-]1 [I-]1 [OH-]1 6. 1.8 x 10-4 = k [0.0010] [0.0030] [1.00] k = 1.8 x 10-4 / [0.0010] [0.0030] [1.00] = 60

For the hypothetical reaction: A + B → C + D we can write the following rate law with hypothetical exponents which need to be determined. Rate = k [A]x [B]y

Rate = k [A]x [B]y f we run separate experiments with varying concentrations which yield relative rates (rate 1 and rate 2) we can write the following equation to compare these rates: rate 1 = k[A1]x [B1]x ___________________________ rate 2 = k [A2]x B2]x

Ex.

Rate = k [ClO-]x [I-]y [OH-]z 1.8 x 10-4 = k [0.0010]x [0.0030]y [1.00]z ___________________________________ 3.6 x 10-4 = k [0.0020]x [0.0030]y [1.00] z 1.8 x 10-4 = k [0.0010]x _______ ___________ 3.6 x 10-4 = k[0.0020]x 0.50 = (0.5)x x= 1 with respect to ClO- 3.6 x 10-4 = k [0.0020]x [0.0030] y[1.00]z ___________________________________ 7.2 x 10-4 = k [0.0020]x [0.0060]y [1.00] z 3.6 x 10-4 = k [0.0030]y _______ ___________ 7.2 x 10-4 = k[0.0060]y (0.5) = (0.5) y y= 1 with respect to I- 1.8 x 10-4 = k[0.0010]x [0.0030]y [1.00]z __________________________________ 9.0 x 10-5 = k[0.0010]x [0.0030]y[0.50]z 1.8 x 10-4 = k [1.00]z ___________________ 9.0 x 10-5 = k [0.50]z 2.0 = (2.0)z z= 1 with respect to OH- The overall rate can now be written as k [ClO-]1 [I-]1 [OH-]1 the rate constant k= 1.8 x 10-4 = k [0.0010] [0.0030] [1.00] k = 1.8 x 10-4 / [0.0010] [0.0030] [1.00] = 60

First-Order Reaction

Reactions are most commonly first order with respect to most reactants. For a first-order reaction, the rate of reaction is directly proportional to the concentration of one of the reactants. The most common example of this type of reaction is the radioactive fission of unstable nuclei such as 238U92 as shown below: 238U92 → 234Th90 + 4He2 Rate Law: r = k [A] the rate constant, k, has units of 1/sec.

The instantaneous rate near the end of the reaction will be __________ than the instantaneous rate at the end of the reaction or the average rate.

Smaller

Collision Theory

The dependency of reaction rates on concentration and temperature is explained in part by the collision theory which states that for a reaction to occur, particles must collide with sufficient energy (activation energy, Ea) and proper alignment. Therefore, the rate depends on three factors, (1) how often particles collide, (2) how many of these collisions have the proper alignment and (3) how many of these collisions have energy greater than the required activation energy

The order of the reaction

The exponent of each concentration in the rate law is called the order of the reaction with respect to that reactant

If given this, how to do it

The following rate data was obtained for the reaction which takes place in a solution of OH-: ClO- + I- → IO- + Cl- the following data table is obtained: Experiment # [I-] [ClO-] [OH-] rate 1 0.0030 0.0010 1.00 1.8 x 10-4 2 0.0030 0.0020 1.00 3.6 x 10-4 3 0.0060 0.0020 1.00 7.2 x 10-4 4 0.0030 0.0010 0.50 9.0 x 10-5 Determine the reaction order with respect to (1) ClO-, (2) I- and (3) OH-, (4) write the rate law and then (5) determine the value of the rate constant, k.

The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of predominately reactants, predominately products or significant amounts of both products and reactants. Be sure to explain your answer. N2 (g) + 3 H2 (g) <==> 2NH3 (g) Kc= 4.1 x 10>8

The very large Kc indicates that this equilibrium mixture will be composed of mostly products.

The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of predominately reactants, predominately products or significant amounts of both products and reactants. Be sure to explain your answer. N2 (g) + O2 (g) <==> 2 NO (g) Kc= 4.6 x 10>-31

The very small Kc indicates that this equilibrium mixture will be composed of mostly reactants.

chemical equilibrium

These reactions usually reach a point where the reaction mixture ceases to change in concentration of any component since the forward and reverse reactions are occurring at the same rates at which time the reaction is said to be in chemical equilibrium.

The Kc value can also be used to determine the equilibrium concentrations of the reactants and products if the starting concentrations of the reactants are known.

This is very useful because usually you begin a reaction with known quantities of reactants and want to know what amounts of materials will be present once the reaction is finished (in the case of these reaction, when the reaction has come to equilibrium). Let's use the following example to demonstrate this process.

Natural logs (ln) can be determined on a scientific calculator.

To determine the natural log of a number, enter the number and push the "ln" button. To carry out the anti-natural log operation, enter the natural log value and push "2nd" and then "ln" button.

In the reaction of 0.0200 M gaseous N2O5 to yield NO2 gas and O2 gas as shown below: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) This will decompose to form this

Using the data from the table, we can calculate the average rate over the measured time interval from 0 to 700 secs as follows using the [N2O5] or [O2] (or [NO2] if we wanted): rate = Δ[O2] / Δt = (0.0070- 0) / 700 - 0 = 1.00 x 10-5 mol/L•s rate = - Δ[ N2O5] / Δt = - (0.0060 - 0.0200) / 700 - 0 = 2.00 x 10-5 mol/L•s

How to calculate Kc: EX If we study the catalytic methanation reaction and look at actual reaction conditions, we find that when 1.000 mole of CO and 3.000 mole of H2 are placed in a 10.00 L container at 927°C along with a catalyst and allowed to come to equilibrium, they form an equilibrium mixture of 0.613 mole of CO, 1.839 mole of H2, 0.387 mole of CH4 and 0.387 mole of H2O.

What occurred: 0.387 mole of CO and 1.161 mole of H2 have reacted to give 0.387 mole of CH4 and 0.387 mole of H2O

Equilibrium constant Kp

When discussing equilibrium reactions which involve only gases (like catalytic methanation), it is usually easier to express the equilibrium constant in terms of partial pressures of the gases since partial pressures are easier to determine than concentrations, and concentration is directly proportional to the partial pressures (at a set temperature) since n/V (moles/L) = P/RT from the ideal gas law. The equilibrium constant is designated as Kp when partial pressures (P) are used instead of mole/L concentrations and its numerical value is different than that of Kc. So the Kp for catalytic methanation would be: Kp= Pch4 Ph20 _______________ PCo PH2>3

heterogeneous equilibrium

When reactants and products are present in more than one phase(s, g, l etc), the equilibrium is called a heterogeneous equilibrium and several examples are shown below: 4 H2 (g) + Fe3O4 (s) <==> 3 Fe g(s) + 4 H20 (g) 4 H20 (liq) + 2 Cl2 (g) <==> 4 HCl (aq) + O2(g)

Transition state

When the reactant molecules are provided with energy equal to the activation energy, they combine to form the transition state, an unstable complex of atoms which can decompose to form the products and example of which is shown below:

Explanation of how transition state works

When the reactants come together with sufficient energy and in proper alignment, a bond begins to form between the nitrogen atom of the NO molecule and one of the oxygen atoms of the O3 molecule while at the same time the bond between that oxygen atom of the O3 molecule begins to break. This unstable complex of atoms (with bonds forming and breaking) is the transition state. Once the bond fully forms between the nitrogen atom of the NO molecule and the oxygen atom of the O3 molecule and the bond fully breaks between that oxygen atom and the rest of the O3 molecule, the products have formed.

Second-Order Rate Law For a second-order reaction, the rate law with respect to time is:

1/[A] - 1/ [A0] = kt where [A]0 is initial concentration, [A] is concentration at time t and k is the rate constant. This type of reaction order is uncommon enough that we will not deal with these type reactions in this course.

From this you can now substitute concentration data and rate data for any of the experiments and determine the rate constant, k, as follows using data from experiment 1:

3.0 x 10-5 = k [0.100]1 [0.200]2 3.0 x 10-5 = k [0.100] [0.040] k = 3.0 x 10-5 / 0.0040 = 7.5 x 10-3

ANytime theres a liquid or solid, the concentration doesnt change (constant) so their concentration does not appear in the equation

4 H20 (liq) + 2 Cl2 (g) <==> 4 HCl (aq) + O2(g) Drop out the liq

The equilibrium reaction below has the Kc = 4.00 at 25°C. If the temperature of the system at equilibrium is increased to 100°C, how and for what reason will the equilibrium shift. Also show and explain how and why the Kc value will change. 2 SO3 <==> 2 SO2 + O2 ChangeH = +94.1 kJ

If the temperature is increased on this reaction at equilibrium which has a +ΔH0, the reaction will briefly shift in the direction that uses up some of the added heat [+ΔH0 indicates an endothermic (heat absorbing) forward reaction] so this reaction will shift in the forward direction. Simultaneously, this forward shift will increase the concentration of SO2 and O2 (in the numerator of Kc) and decrease the concentration of SO3 (in the denominator of Kc) which will increase the value of Kc.

The equilibrium reaction below has the following equilibrium mixture concentrations: SO3 = [0.0894], SO2 = [0.400] and O2 = [0.200] with Kc = 4.00 If the concentration of O2 at equilibrium is decreased to [0.100], how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2SO3 (g) <==> 2 SO2 + O2

Kc = 4.00 when SO3 = [0.0894], SO2 = [0.400] and O2 = [0.200] TO find Qc: Q = [0.400]2 [0.100] ----------------- [0.0894]2 Qc=2 which is less than the Kc The reaction must shift briefly in the forward direction to increase the [O2] to come back to equilibrium. This is in agreement with Qc < Kc which also predicts the reaction will proceed to the right.

The equilibrium reaction below has the following equilibrium mixture concentrations: SO3 = [0.0894], SO2 = [0.400] and O2 = [0.200] with Kc = 4.00 If the concentration of SO3 at equilibrium is increased to [0.300], how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2 SO3 <==> 2 SO2 + O2

Kc = 4.00 when SO3 = [0.0894], SO2 = [0.400] and O2 = [0.200] Q= [0.400]2 [0.200] ---------------------------- = [0.300]2 Q = 0.032/ 0.09 = 0.356 The reaction must shift briefly in the forward direction to decrease the [SO3] to come back to equilibrium. This is in agreement with Qc < Kc which also predicts the reaction will proceed to the right.

Effect of Solvent

Many reactions take place in solution and the solvent used to dissolve the reactants can have remarkable effects on not only the rate of reaction but what products are formed and how much of each product is formed. Solution reactions are generally faster than those in heterogeneous mixtures since the reactants can come in contact with one another much more freely in solution. Reactions in heterogeneous mixtures occur only where the phases contact one another and these heterogeneous reactions can be speeded up by stirring which increases the amount of contact or by reducing the particle size of the reactants which increases their surface area by which they come in contact.

Effect of Temperature changes

Most reactions are markedly affected by temperature since reaction rates generally increase as the temperature is increased due to greater kinetic energy of the reacting particles which will allow a reaction to come to equilibrium more quickly. But temperature also affects the value of the equilibrium constant, Kc (or Kp), the numerical value of Kc (or Kp) is only constant for a given reaction at a certain temperature (the value of Kc (or Kp) is NOT affected by concentration changes, pressure change or addition of catalysts). To predict how to increase the rate of the forward reaction so as to increase the yield of products for an equilibrium reaction, one must know whether the forward reaction is endothermic (+ΔH0) or exothermic (-ΔH0). The value of ΔH0 for catalytic methanation illustrates that the forward reaction is rather exothermic as shown below: CO (g) + 3 H2 (g) <==> CH4 (g) + H2O (g) Temp = -2-6.2 kJ

The magnitude of the Kc value can be used to determine if an equilibrium mixture contains a greater concentration of products or reactants. If the Kc value is large, the equilibrium mixture will contain a higher concentration of products but if the Kc value is small, the equilibrium mixture will contain a higher concentration of reactants. If the value of Kc is neither large nor small but ≈ 1, the equilibrium mixture will contain a significant concentration of both products and reactants. This principle is illustrated with the 3 examples below:

N2 (g) + 3 H2 (g) <==> 2 Nh3 (g) Kc= 4.1 x 10>8 Since the Kc is large, the equilibrium must be predominantly products.

If the equilibrium mixture contains 0.230 mole of N2 then the following amounts of materials must be present in the equilibrium mixture:

N2 = 0.230 mole of (as stated) H2O = 0.230 x 6/2 mole (6 mole of H2O forms for every 2 mole of N2 that is formed) NH3 = 0.700 - 4/2 x 0.230 mole (4 mole of NH3 reacts for every 2 mole of N2 that is formed) O2 = 0.910 - 3/2 x 0.230 mole (3 mole of O2 reacts for every 2 mole of N2 that is formed) Change all amounts to moles/L before entering in Kc expression: N2 = 0.230 mole / 1.00 L = 0.230 M H2O = 0.690 mole / 1.00 L = 0.690 M NH3 = 0.240 mole / 1.00 L = 0.240 M O2 = 0.565 mole / 1.00 L = 0.565 M [0.230]2 [0.690]6 ____________ [0.240]4 [0.565]3 [0.0529] [0.1079] 0.00570791 _____________ __________ [0.0033] [0.1804] 0.00059532 Kc= 9.54

EX: Nitric oxide, NO, is formed by the reaction of N2 and O2 at 2127°C as shown below. Initially a reaction contains 1.00 mole of N2 and 1.00 mole of O2 in a 10.0 L container at 2127°C. How many moles of each material (N2, O2 and NO) will be present in the equilibrium mixture? The value of Kc is 0.0025 at this temperature.

The initial (mol/L) concentration of N2 is 1.00 mol/10.0 L (= 0.100 M) and the initial (mol/L) concentration of O2 is 1.00 mol/10.0 L (= 0.100 M). The initial concentration of NO is 0. As the reaction proceeds to equilibrium, some unknown amount (x) of N2 reacts with some unknown amount (x) of O2 to form some unknown amount (x) of NO. The initial and final (M) concentration of each substance is shown in the chart below. Conc (M) N2 O2 2NO Initial 0.100 0.100 0 Change -x -x +2x Final (equi)0.100 - x 0.100 - x 0 + 2x If you now substitute the equilibrium concentration given for each substance in the equilibrium expression, you get the following:

Half-Life

The rate law shows the relationship between rate and concentrations of reactants but often it would be more useful to have an equation which shows how reactant concentration varies over a period of time. This equation would be more comparable to the actual data which shows concentrations present at different times and would allow determination of a concentration at any particular time or prediction of time required to reach a particular concentration. These equations relating concentration and time are derived by use of calculus and vary depending on the order of the reaction.

Equilibrium Constant Kc EX CO (g) + 3H2 (g) <===> Ch4 (g) + H2O(g)

The rate of the forward reaction (CO + 3 H2) can be described by the following rate law: Ratef = kf [CO] [H2]3 And the rate of the reverse reaction (CH4 + H2O) can be described by the following rate law Rater = kr [CH4] [H2O] Since these rates are equal at equilibrium we can write the following equations: Ratef = Rater or kf [CO] [H2]3 = kr [CH4] [H2O] Rearranging the equation gives us: kf= [CH4] [H20] ______________________ Kr [CO] [H2]3 Since kf / kr (the ratio of two constants) is also a constant, which we will call Kc, for the equilibrium constant, we can write: [CH4] [H20] Kc= _________________ [CO] [H2]3

Order of rate = k [Br-] [BrO3-] [H+]2

The reaction is fourth order overall, first order with respect to Br- and BrO3- and second order with respect to H+.

The equilibrium reaction below has the Kc = 4.00. If the volume of the system at equilibrium is decreased from 4.00 liters to 2.00 liters, how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2 SO3 <==> 2 SO2 + O2

When volume decreases from 4.00 to 2.00, the pressure doubles and the concentration of all gases (SO3, SO2 and O2) doubles so: At equilibrium Qc = Kc = [SO2]2 [O2] / [SO3]2 = 4.0 (volume halved = pressure doubled = conc doubled) Qc = [2SO2]2 [2O2] / [2 SO3]2 = 2Kc The reaction must shift briefly in the direction that decreases the pressure by going toward the side with the lesser moles of gas (reverse direction : 3 moles of gas yields 2 moles of gas) to come This is in agreement with Qc > Kc: the reaction will proceed to the left (in the direction of the reactants).

The equilibrium reaction below has the Kc = 4.00. If the volume of the system at equilibrium is increased from 3.00 liters to 9.00 liters, how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2 SO3 <==> 2 SO2 + O2

When volume increases from 3.00 to 9.00, the pressure is cut to 1/3 of original and the concentration of all gases (SO3, SO2 and O2) decreases to 1/3 so: (at equilibrium) At equilibrium Qc = Kc = [SO2]2 [O2] / [SO3]2 = 4.0 (volume tripled = Pi / 3 = conc / 3) Qc = [1/3 SO2]2 [1/3 O2] / [1/3 SO3]2 = Kc/3 The reaction must shift briefly in the direction that increases the pressure by going toward the side with the greater moles of gas (forward direction : 2 moles of gas yields 3 moles of gas) to come back to equilibrium. This is in agreement with Qc < Kc: the reaction will proceed to the right (in the direction of the products).

Use the [O2] data from the table to calculate the instantaneous rate late in the reaction (2400 secs to 3000 secs).

Will be smaller rate value Answer: rate = Δ[O2] / Δt = (0.054 - 0.052) / 3000 - 2400 = 3.33 x 10-6 mol/L•s

Transition state with catalyst vs without catalyst

With- A catalyst speeds up the rate of a chemical reaction by providing an alternate transition state by which the reaction can occur which has a lower (Eact) energy of activation, sometimes allowing a very slow nonspontaneous reaction to became a faster (sometimes spontaneous) reaction. Without-opposite occurs

We must now convert these equilibrium mole quantities to M / L as shown below and then substitute the quantities in their proper places in the Kc expression being sure to raise any that require it to a power equal to their reaction coefficient.

[CO] = 0.613 mole / 10.00 L = 0.0613 M [H2] = 1.839 mole / 10.00 L = 0.1839 M [CH4] = 0.387 mole / 10.00 L = 0.0387 M [H2O] = 0.387 mole / 10.00 L = 0.0387 M Plug in: Kc= [0.0387] [0.0387] ______________________________ [0.0613] [0.1839]3 Kc= 3.93

catalyst

a substance that increases the rate of a chemical reaction without being consumed in the overall reaction

Since this reaction is not at equilibrium, the normal ratio of products over reactants (shown below) cannot be set equal to the Kc but is instead shown as Qc, the reaction quotient

an expression which has the same form as the equilibrium constant expression but whose concentrations are not equal to those at equilibrium. Qc= [CH4] [H2O] --------------------- [CO] [H2]3 If we put the non-equilibrium concentrations given above into the Qc expressions we get a value of 1.5625 as shown below: Insert given values Compare Qc to Kc