Chem 2 Test 1

A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.)

-8.00 g H3PO4 × (1 mol / 97.994 g H3PO4) = 0.0816 mol H3PO4 -100.0 mL H2O × (1.00 g H2O / mL) × (1 kg / 1000 g) = 0.1000 kg H2O -Molality = 0.0816 mol H3PO4 / 0.1000 kg H2O] = 0.816 m

Reaction Quotient, Q

-a quotient obtained by applying the law of mass action to initial concentrations rather than to equilibrium concentrations -Used when all of the initial concentrations are nonzero. Q = K; The system is at equilibrium. No shift will occur. Q > K; The system shifts to the left. Consuming products and forming reactants, until equilibrium is achieved. Q < K; The system shifts to the right. Consuming reactants and forming products, to attain equilibrium.

Formation of a Liquid Solution

1. Separating the solute into its individual components (expanding the solute). 2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). 3. Allowing the solute and solvent to interact to form the solution.

The Extent of a Reaction

A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products - the equilibrium lies to the right. Reaction goes essentially to completion. A very small value of K means that the system at equilibrium will consist of mostly reactants - the equilibrium position is far to the left. Reaction does not occur to any significant extent.

The value of the rate constant for a gas phase reaction can be changed by increasing the A. pressure of the reactant. B. amount of product. C. temperature of the reaction vessel. D. volume of the reaction vessel.

According to Arrehenius Equation , K = Ae^(-Ea /RT) Where K = rate constant T = temperature R = gas constant Ea = activation energy A = Frequency factor(constant) So rate constant can be increased by changing thetemperature. So option C is correct

Temperature Effects (for Aqueous Solutions)

Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature. Predicting temperature dependence of solubility is very difficult. Solubility of a gas in solvent typically decreases with increasing temperature.

The chemicals in a lightstick give off light as they react. When the lightstick is placed in warm water the glow increases. This is because the a. higher temperature changes the wavelength of light emitted. b. average kinetic energy of the reactants increases. c. activation energy for the process is lowered. d. higher temperature catalyzes the reaction.

B. due to the heat of the water

Which of the following statements is FALSE? A. In order for a reaction to occur, reactant molecules must collide. B. A catalyst alters the rate of a reaction and is neither a product nor a reactant in the overall equation. C. A reaction of the form, 2A+B----products, will usually have a one step mechanism. D. The rate of the bimolecular reaction A+B-----products is proportional to the freuency of collisions between A and B. E. The transition state is a short lived, high energy state, intermediate between reactants and products.

C

Reaction Rate

Change in concentration of a reactant or product per unit time.

A 1.00 mol sample of N2O4(g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation: N2O4(g)⇌2NO2(g) K = 4.00 × 10-4 Calculate the equilibrium concentrations of: N2O4(g) and NO2(g).

Concentration of N2O4 = 0.097 M Concentration of NO2 = 6.32 x 10-3 M (without quadratic) or Concentration of N2O4 = 0.094 M Concentration of NO2 = 6.22 x 10-3 M (with quadratic) Use this problem to discuss the 5% allowable error (so we can assume x is negligible). Make sure the students understand we are NOT saying x is equal to zero but that x is negligible. This is a subtle but very important point.

Le Chatelier's Principle Effects of Changes on the System

Concentration: The system will shift away from the added component. If a component is removed, the opposite effect occurs. Temperature: K will change depending upon the temperature (endothermic - energy is a reactant; exothermic - energy is a product). Pressure: a.) The system will shift away from the added gaseous component. If a component is removed, the opposite effect occurs. b.) Addition of inert(noble) gas does not affect the equilibrium position. c.) Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas.

The effect of temperature on the rates of chemical reactions is primarily a result of the (A) size of the colliding molecules. (B) orientation of the colliding molecules. (C) enthalpies of the reactants and products. (D) kinetic energies of the colliding molecules.

D

Colligative Properties

Depend only on the number (concentration), not on the identity, of the solute particles in an ideal solution: Boiling-point elevation Freezing-point depression Osmotic pressure

Coagulation

Destruction of a colloid. Usually accomplished either by heating or by adding an electrolyte.

Types of Rate Laws

Differential Rate Law (rate law) - shows how the rate of a reaction depends on concentrations, i.e. [A] in mol/L Integrated Rate Law - shows how the concentrations of species in the reaction depend on time, i.e. deltat. Experimental convenience usually dictates which type of rate law is determined experimentally. Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law.

Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25°C to 35°C?

Ea = 53 kJ ln(2) = (Ea / 8.3145 J/K·mol)[(1/298 K) - (1/308 K)]

Conclusions About the Equilibrium Expression

Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When the balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus Knew = (Koriginal)^n. K values are usually written without units. K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there are many equilibrium positions but only one equilibrium constant, K. Equilibrium position is a set of equilibrium concentrations.

Consider the reaction represented by the equation: Fe3+(aq) + SCN-(aq) FeSCN2+(aq) Trial #1: 6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN2+(aq) is 4.00 M. What is the value for the equilibrium constant for this reaction?

Fe3+(aq) + SCN-(aq)⇌FeSCN2+(aq) Initial 6.00 10.00 0.00 Change - 4.00 - 4.00 +4.00 Equilibrium 2.00 6.00 4.00 K= ([FeSCN²⁺]/[Fe³⁺][SCN⁻])= ([4.00M]/[2.00M][6.00]) K = 0.333

Chemical Equilibrium

In a chemical reaction, the state in which the rate of the forward reaction equals the rate of the reverse reaction, so that the relative concentrations of the reactants and products do not change with time.

A 2.0 mol sample of ammonia is introduced into a 1.00 L container. At a certain temperature, the ammonia partially dissociates according to the equation: NH3(g) ⇌N2(g) + H2(g) At equilibrium 1.00 mol of ammonia remains. Calculate the value for K.

K = 1.69 This answer assumes the students have balanced the equation with relative coefficients of 2:1:3.

A solution was prepared by dissolving 25.00 g of glucose in 200.0 g water. The molar mass of glucose is 180.16 g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution. Kb = 0.51 °C·kg/mol.

Kg of solvent = (200.0 g)(1 kg / 1000 g) = 0.2000 kg water msolute = (0.1388 mol glucose) / (0.2000 kg water) = 0.6938 mol/kg ΔT = (0.51 °C·kg/mol)(0.6938 mol/kg) = 0.35 °C. The boiling point of the resulting solution is 100.00 °C + 0.35 °C = 100.35 °C.

The Relationship Between K and Kp

Kp = K(RT)Δn Δn = sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. R = 0.08206 L·atm/mol·K T = temperature (in Kelvin)

N2(g) + 3H2(g) ⇌2NH3(g) Using the value of Kp (3.9 × 104) from the previous example, calculate the value of K at 35°C.

Kp= K(RT)^(∆ⁿ) 3.9x10⁴=K(0.08206 L∙atm/mol∙k x 308K)(²⁻⁴) K=2.5x10⁷

Nonideal Solutions

Liquid-liquid solutions where both components are volatile. Modified Raoult's Law: Nonideal solutions behave ideally as the mole fractions approach 0 and 1.

Pressure Effects on Solubility

Little effect on solubility of solids or liquids Henry's law: C = kP C = concentration of dissolved gas k = constant P = partial pressure of gas solute above the solution Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.

Boiling-Point Elevation

Nonvolatile solute elevates the boiling point of the solvent. ΔT = Kbmsolute ΔT = boiling-point elevation Kb = molal boiling-point elevation constant msolute = molality of solute

Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role.

Oil is a mixture of nonpolar molecules that interact through London dispersion forces, which depend on molecule size. ΔH1 will be relatively large for the large oil molecules. The term ΔH3 will be small, since interactions between the nonpolar solute molecules and the polar water molecules will be negligible. However, ΔH2 will be large and positive because it takes considerable energy to overcome the hydrogen bonding forces among the water molecules to expand the solvent. Thus ΔHsoln will be large and positive because of the ΔH1 and ΔH2 terms. Since a large amount of energy would have to be expended to form an oil-water solution, this process does not occur to any appreciable extent.

The Energy Terms for Various Types of Solutes and Solvents

One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed. Processes that require large amounts of energy tend not to occur. Overall, remember that "like dissolves like".

Raoult's Law:

Psoln=XsolvP°solv Psoln =observed vapor pressure of solution Xsolv=mole fraction of solvent P°solv=vapor pressure of pure solvent

First Order rate law equation

Rate = k[A] Integrated: ln[A] = -kt + ln[A]o [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A

second order rate law

Rate=k[A]^2

Le Chatelier's Principle

States that if a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

Steps in the Dissolving Process

Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent. Step 3 usually releases energy. Steps 1 and 2 are endothermic, and step 3 is often exothermic. Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔHsoln = ΔH1 + ΔH2 + ΔH3 ΔHsoln may have a positive sign (energy absorbed) or a negative sign (energy released).

Consider an equilibrium mixture in a closed vessel reacting according to the equation: H2O(g) + CO(g) ⇌ H2(g) + CO2(g) You add more H2O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

The concentrations of each product will increase, the concentration of CO will decrease, and the concentration of water will be higher than the original equilibrium concentration, but lower than the initial total amount. Students may have many different answers (hydrogen goes up, but carbon dioxide in unchanged, etc.) Let them talk about this for a while - do not go over the answer until each group of students has come up with an explanation. This question also sets up LeChâtelier's principle for later.

Consider an equilibrium mixture in a closed vessel reacting according to the equation: H2O(g) + CO(g) ⇌ H2(g) + CO2(g) You add more H2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

The concentrations of water and CO will increase. The concentration of carbon dioxide decreases and the concentration of hydrogen will be higher than the original equilibrium concentration, but lower than the initial total amount.

Van't Hoff Factor, i

The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur). NaCl i = 2 KNO3 i = 2 Na3PO4 i =4

Enthalpy (Heat) of Solution

The net energy change during the dissolving process

Instantaneous Rate

The rate of decomposition at a specific time, calculated from the rate law, the specific rate constant, and the concentrations of all the reactants.

Overall Reaction Order

The sum of the exponents in the reaction rate equation. Rate = k[A]^n[B]^m Overall reaction order = n + m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B Remember, n and m do NOT relate to the stoichiometric coefficients in the balanced chemical reaction.

Method of Initial Rates

The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible. Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants.

Half life equation for first order

Time required for a reactant to reach half its original concentration k = rate constant Half-life does not depend on the concentration of reactants.

Rate Law

an expression for the rate of a reaction in terms of the concentration of reactants. For the decomposition of nitrogen dioxide: 2NO2(g) → 2NO(g) + O2(g) [Reaction is balanced!] Rate = k[NO2]^n: k = rate constant n = order of the reactant The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate. The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation.

1) According to the collision theory of kinetics, which statement best describes the rate of a chemical reaction? a. The greater the difference in energy between the reactants and products, the faster is the reaction b. All collisions between molecules with at least a minimum kinetic energy result in reaction. c. All collisions between molecules with at least a minimum kinetic energy and the proper orientation result in reaction. d. The greater the difference in energy between the reactant and transition state, the faster is the reaction. e. All collisions result in a chemical reaction.

c. All collisions between molecules with at least a minimum kinetic energy and the proper orientation result in reaction.

Half life equation for second order

gets longer as the reaction progresses and the concentration of reactants decrease Each successive half-life is double the preceding one

Heterogeneous equilibria

involves more than on phase (solid, liquid, gas) ; position of equilibrium does not depend on the amounts of solids or liquids present - because the concentrations of pure solids and liquids cannot change 2KClO3(s)⇌2KCl(s) + 3O2(g) K=[O₂]³

A first order reaction is 35% complete at the end of 55 minutes. What is the value of k?

ln(0.65) = -k(55) + ln(1)

Mass (weight) percent

mass of solute/mass of solution x 100%

Molality (m)

moles of solute/kg of solvent

Molarity (M) equation

moles of solute/liters of solution

mole fraction equation

moles of solute/moles of solute + moles of solvent

Factors affecting solubility

nature of solute and solvent, temperature, pressure Structure Effects: Polarity Hydrophobic (water fearing) -Non-polar substances Hydrophilic (water loving) -Polar substances Pressure Effects: Henry's law Temperature Effects: Affecting aqueous solutions

osmotic pressure

pressure that must be applied to prevent osmotic movement across a selectively permeable membrane

The reaction A + 2B → C has the following proposed mechanism: A + B ⇌ D (fast equilibrium) D + B → C (slow) Write the rate law for this mechanism.

rate = k[A][B]2

If the equilibrium lies to the right, the value for K is __________. If the equilibrium lies to the left, the value for K is ___________.

right= large (or >1) left= small (or <1)

nonvolatile solute

the vapor pressure of a solvent containing a _______ is lower than the vapor pressure of the pure solvent at the same temperature It serves to lowers the vapor pressure of a solvent

Arrhenius Equation

to find activation energy use the... A = frequency factor Ea = activation energy R = gas constant (8.3145 J/K·mol) T = temperature (in K)

For each of the following solutions, would you expect it to be relatively ideal (with respect to Raoult's Law), show a positive deviation, or show a negative deviation? a) Hexane (C6H14) and chloroform (CHCl3) b) Ethyl alcohol (C2H5OH) and water c) Hexane (C6H14) and octane (C8H18)

a) Positive deviation; Hexane is non-polar, chloroform is polar. b) Negative deviation; Both are polar, and the ethyl alcohol molecules can form stronger hydrogen bonding with the water molecules than it can with other alcohol molecules. c) Ideal; Both are non-polar with similar molar masses.

For a reaction aA→ Products, [A]₀ = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. Write the rate law for this reaction. b) Calculate k. Calculate [A] at t = 525 minutes.

a) rate = k[A]2 We know this is second order because the second half-life is double the preceding one. b) k = 8.0 x 10-3 M-1min-1 25 min = 1 / k(5.0 M) c) [A] = 0.23 M (1 / [A]) = (8.0 x 10-3 M-1min-1)(525 min) + (1 / 5.0 M)

Consider the reaction represented by the equation: Fe3+(aq) + SCN⁻(aq)⇌ FeSCN2+(aq) Fe3+ SCN- FeSCN2+ Trial #1 9.00 M 5.00 M 1.00 M Trial #2 3.00 M 2.00 M 5.00 M Trial #3 2.00 M 9.00 M 6.00 M Find the equilibrium concentrations for all species.

Trial #1: [Fe3+] = 6.00 M; [SCN-] = 2.00 M; [FeSCN2+] = 4.00 M Trial #2: [Fe3+] = 4.00 M; [SCN-] = 3.00 M; [FeSCN2+] = 4.00 M Trial #3: [Fe3+] = 2.00 M; [SCN-] = 9.00 M; [FeSCN2+] = 6.00 M This problem will provide a good discussion of Q vs. K. Trial #1 proceeds to the right to reach equilibrium, Trial #2 proceeds to the left, and Trial #3 is at equilibrium. Watch for students setting up an ICE chart without thinking about which direction the reaction must proceed initially. Be prepared for some discussion about the fact that in the Change row we can have a "-x" on the right side and a "+x" on the left side and still use the same expression for K. Use this so that students can think about the direction the reaction must proceed initially so that they needed memorize a relationship between Q and K. The students can solve this with the quadratic but it is not necessary. The numbers have been chosen to be relatively easy to solve. Tell the students the problems will not always have numbers like these, so take the time to understand what is going on (so the math doesn't "get in the way").

Elementary Steps (Molecularity)

Unimolecular - reaction involving one molecule; first order. Bimolecular - reaction involving the collision of two species; second order. Termolecular - reaction involving the collision of three species; third order. Very rare.

Freezing-Point Depression

When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. ΔT = Kfmsolute ΔT = freezing-point depression Kf = molal freezing-point depression constant msolute = molality of solute

Solving Equilibrium Problems

Write the balanced equation for the reaction. Write the equilibrium expression using the law of mass action. List the initial concentrations. Calculate Q, and determine the direction of the shift to equilibrium. Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown. Check your calculated equilibrium concentrations by making sure they give the correct value of K.

Consider the reaction aA → Products. [A]₀ = 5.0 M and k = 1.0 × 10-2 (assume the units are appropriate for each case). Calculate [A] for each order after 30.0 seconds have passed, assuming the reaction is

Zero order-4.7 M First order -3.7 M Second order-2.0