Chapter 14 B

Consider the following multi-step reaction: A+B→AB(slow) A+AB→A2B(fast) −−−−−−−−−−−−−−−− 2A+B→A2B(overall) Based on this mechanism, determine the rate law for the overall reaction.

rate = k [A][B] ; the rate law is always based on the slow reaction

Consider the following multi-step reaction: C+D⇌CD(fast) CD+D→CD2(slow) CD2+D→CD3(fast) −−−−−−−−−−−−−−−−− C+3D→CD3(overall) Based on this mechanism, determine the rate law for the overall reaction.

rate = k [C][D]^2

Activation Energy

the minimum amount of energy required to start a chemical reaction

The following mechanism has been proposed for the gas-phase reaction of H2 with ICl: H2(g)+ICl(g)→HI(g)+HCl(g) HI(g)+ICl(g)→I2(g)+HCl(g) Identify any intermediates

HI

Rate Constant (k)

Initial Rate (M/s)/Reactant Concentrations (M)

Given the following diagrams at t = 0 ([A] = 8) and t = 30 ([A] = 2) What is the half-life of the reaction if it follows first-order kinetics?

* t 1/2 = 0.693/k * FIND k : -kt = ln (af 2/ai 8) = -k(30 min) = (-1.3863/-30) = k k = 0.0462 min -1 FIND half life: t1/2 = 0.693/0.0462 = 15 Min

Homogeneous Catalyst

- Catalyst present in the same phase as reactants

Catalyst

- Substance that speeds the rate of a chemical reaction without undergoing a permanent chemical change - Present at the start of a reaction

A first-order reaction A⟶B has the rate constant k= 4.2×10−3 s−1 . If the initial concentration of A is 2.8×10−2 M, what is the rate of the reaction at t= 620 s ?

1) * ln[A] = -kt + ln [A]o * 2) Rate = k [A] * Given k and [A]0, use the integrated form of the 1st order rate law to calc the [A] at t = 620s; then rate = k[A] ln [A] = (-4.2x10-3 s-1) (620s) + ln (2.8x10-2 M) Ln [A] = -6.1796 e^-6.1796 [A] = 2.07x10-3 M ** rate = k [A] = (4.2x10-3 s-1) x (2.07x10-3 M) Rate = 8.7x10-6 M/s

Termolecular Reaction Rate Laws

A + A + A ---> products = k [A]^3 A + A + B ---> products = k [A]^2[B] A + B + C ---> products = k [A][B][C]

Bimolecular Reaction Rate Laws

A + A ---> products = k [A]^2 A + B ---> products = k[A][B]

Unimolecular Reaction Rate Law

A --> products = k[A]

The accompanying graph shows plots of ln k versus 1/T for two different reactions. The plots have been extrapolated to the y-intercepts. (Figure 1) Red = Top, negative Slope Blue Bottom, negative Slope Which reaction has the larger value for Ea?

Blue; the magnitude of the slope is greater than that of the red line

Average Rate of Appearance

Change in Concentration of Products (M)/Change in Time (s) Rate = Delta [Products]/Delta time

Average Rate of Disappearance

Change in Concentration of Reactants (M)/Change in Time Rate = -Delta [Reactants]/Delta Time

The gas-phase reaction Cl(g)+HBr(g)⟶HCl(g)+Br(g) has an overall energy change of -66 kJ. The activation energy for the reaction is 7 kJ. What's the Activation Energy for the reverse reaction?

Ea-reverse = 66kJ+7kJ = 73 kJ

Using the following kinetic data, determine the magnitude of the first-order rate constant: Time (s) Pressure SO2Cl2(atm) 0 1.000 2500 0.947 5000 0.895 7500 0.848 10000 0.803

Find ln SO2CL2 Find Slope (Delta Y/Delta X) - Slope = + k value

The following mechanism has been proposed for the gas-phase reaction of H2 with ICl: H2(g)+ICl(g)→HI(g)+HCl(g) HI(g)+ICl(g)→I2(g)+HCl(g) Write the balanced equation for the overall reaction.

H2(g)+ICl(g)→HI(g)+HCl(g) (+) HI(g)+ICl(g)→I2(g)+HCl(g) ________________________________________ H2(g) + 2ICl(g) ---> I2(g) + 2HCl (g)

In the figure below, we see that Br−(aq) catalyzes the decomposition of H2O2(aq) into H2O(l) and O2(g). Suppose that some KBr(s) is added to an aqueous solution of hydrogen peroxide. Choose the correct sketch of [Br−(aq)] versus time from the addition of the solid to the end of the reaction.

KBr (s) is added to H2O (aq), at t = 0. Assume KBr dissolves instantly. As the reaction proceeds the catalyst is consumed and then regenerated. Sketch line begins even, dips down, and then back up to level off at the same as the beginning

Rate

Rate Constant k(M-1/s-1) x Reactant Concentrations (M)

The accompanying graph shows plots of ln k versus 1/T for two different reactions. The plots have been extrapolated to the y-intercepts. (Figure 1) Red = Top, negative Slope Blue Bottom, negative Slope Which reaction has the larger value for the frequency factor, A?

Red; the y-intercept is greater for the red line

If the first step is slow and the second one is fast, what rate law do you expect to be observed for the overall reaction?

The first step's rate law would be used for the overall reaction

Choose the graph showing the reaction pathway for an overall exothermic reaction with two intermediates that are produced at different rates.

The reaction is exothermic because the energy of the products is less than the energy of the reactants Intermediates are dips (2) in this case, and they are formed at different rates

The following mechanism has been proposed for the gas-phase reaction of H2 with ICl: H2(g)+ICl(g)→HI(g)+HCl(g) HI(g)+ICl(g)→I2(g)+HCl(g) Write rate laws for each elementary reaction in the mechanism.

The slow step determines the rate law Rate = k [H2] [ICl]

Does a catalyst lower the overall reaction energy?

Yes, due to providing a different mechanism for the reaction

Given the following diagrams at t = 0 ([A] = 8)and t = 30 ([A] = 2) After four half-life periods for a first-order reaction, what fraction of reactant remains?

[A]t =(.5x.5x.5x.5) = 0.0625 [A] = 8; after one half life = 4 2 half lives = 2 3 half lives = 1 4 half lives = .5

Half Life

t 1/2 = 0.693/k