Online Quiz 4

Consider the reaction 2A + 2B + C → 2D + E If the rate law for this reaction is Rate = k[A][B]^2, what will be the effect on the rate if the concentrations of A, B and C are all doubled at the same time? A. The rate will increase by a factor of 2. B. The rate will increase by a factor of 6. C. The rate will increase by a factor of 8. D. The rate will increase by a factor of 4. E. More information is needed before this question can be answered.

Doubling the concentration of A will double the rate, since it the reaction is first order in A. Doubling the concentration of B will quadruple the rate, since it the reaction is second order in B. Doubling the concentration of C will have no effect on the rate, since it the reaction is zero order in C (i.e. [C] does not appear in the rate law at all). Multiplied together, these effects will increase the rate by a factor of 8: Rate = k (2[A])(2[B])^2 = 2*2^2 k[A][B]^2 = 8 * k[A][B]^2 The correct answer is: The rate will increase by a factor of 8.

For the reaction 3A(g) + 2B(g) → 2C(g) + 2D(g) the following data were collected at constant temperature. Determine the correct rate law for this reaction. Trial Initial [A] Initial [B] Initial Rate (mol/L) (mol/L) (mol/(L·min)) 1 0.200 0.100 6.00 × 10-2 2 0.100 0.100 1.50 × 10-2 3 0.200 0.200 1.20 × 10-1 4 0.300 0.200 2.70 × 10-1

Examine the reaction data for pairs of experiments in which the concentration of one reactant changes while the other stays the same. Between Experiment 2 and Experiment 1, [A] is doubled (0.100 to 0.200) and the Rate is quadrupled (1.50 x 10^-2 to 6.00 x 10^-2). Therefore the reaction is second order in A. Between Experiment 1 and Experiment 3, [B] is doubled (0.100 to 0.200) and the Rate is doubled (6.00 x 10^-2 to 1.2 x 10^-1). Therefore the reaction is first order in B. The complete rate law is: Rate = k[A]^2 [B] The correct answer is: Rate = k[A]^2 [B]

The reaction X → Y is first-order overall and first-order with respect to the reactant X. The result of doubling the initial concentration of X will be to A. shorten the time taken to reach equilibrium. B. shorten the half-life of the reaction. C. increase the rate constant of the reaction. D. decrease the rate constant of the reaction. E. double the initial rate.

For a first-order reaction, the rate is directly proportional to the initial concentration of X. If [X] doubles, the rate will also double: Rate 1 = k[X] Rate 2 = k(2[X]) = 2*k[X] Rate 2 = 2*Rate The correct answer is: double the initial rate.

Which one of the following sets of units is appropriate for a second-order rate constant? A. mol^2 L^-2 s^-1 B. L^2 mol^-2 s^-1 C. L mol^-1 s^-1 D. mol L^-1 s^-1 E. s^-1

For a second-order process, the rate law would be: Rate = k [A]^2 Since the units of rate must be mol L^-1 s^-1, and the units of [A] must be mol L^-1, the units of k must be L mol^-1 s^-1: (L mol^-1 s^-1)*(mol L^-1)^2 = mol L^-1 s^-1 The correct answer is: L mol^-1 s^-1

An increase in temperature increases the reaction rate because A. temperature acts as a catalyst in chemical reactions. B. the activation energy of the reaction will decrease. C. the activation energy of the reaction will increase. D. more collisions will have enough energy to exceed the activation energy. E. a greater fraction of the collisions have the correct orientation of molecules.

Increasing temperature increases the average kinetic energy of molecules, giving more of them enough energy to exceed the activation energy barrier in a collision. The correct answer is: more collisions will have enough energy to exceed the activation energy.

When the reaction A → B + C is studied, a plot of ln[A]subt vs. time gives a straight line with a negative slope. What is the order of the reaction? A. Zero B. First Correct C. Second D. Third E. More information is needed to determine the order.

The first-order integrated rate law is ln([A]sub0/[A]subt) = kt, or, in line format, ln[A]subt = - kt + ln[A]sub0. Therefore a first-order reaction will give a straight line for the plot of ln[A] vs. time, with a slope equal to - k. The correct answer is: First

The decomposition of hydrogen peroxide is a first-order process with a rate constant of 1.06 × 10^-3 min^-1. How long will it take for the concentration of H2O2 to drop from 0.0200 M to 0.0120 M? A. 31,400 min B. 4550 min C. < 1 min D. 7.55 min E. 482 min

The integrated rate law for a first-order process is as follows: ln([A]o/[A]t) = kt If [A]o = 0.0200 M and [A]t = 0.0120 M, then: ln(0.0200/0.0120) = 0.5108 = (1.06 × 10^-3 min^-1)t t = 482 min The correct answer is: 482 min

Given the following balanced equation, determine the rate of reaction with respect to [NOCl]. 2 NO(g) + Cl2(g) → 2 NOCl(g)

The rate is expressed as the change in concentration of reactants/products over time. With respect to [NOCl], that is: Rate = + Δ[NOCl] / 2Δt The rate expression is positive because NOCl is a product, and it is divided by the molar coefficient of 2. The correct answer is: Rate = + Δ[NOCl] / 2Δt

Tetrafluoroethylene, C2F4, can be converted to octafluorocyclobutane which can be used as a refrigerant or an aerosol propellant. A plot of 1/[C2F4] vs. time gives a straight line with a slope of 0.0448 L mol^-1 s^-1. What is the rate law for this reaction?

The second-order integrated rate law is: (1/[A]subt) - (1/[A]sub0) = kt. Therefore, the plot of 1/[A] vs. time would give a straight line with slope equal to k. This reaction must be second order, with k = 0.0448 L mol^-1 s^-1. The rate law is: Rate = 0.0448 (L mol^-1 s^-1)[C2F4]^2 The correct answer is: Rate = 0.0448 (L mol^-1 s^-1)[C2F4]^2

The rate law for the reaction 3A → 2B is rate = k[A] with a rate constant of 0.0447 hr^-1. What is the half-life of the reaction? A. 15.5 hr B. 0.0645 hr C. 44.7 hr D. 22.4 hr E. 0.0224 hr

This is a first-order reaction; therefore the half-life can be calculated according to the formula: t sub 1/2 = ln2 / k = 0.6931 / 0.0447 hr^-1 t sub 1/2 = 15.5 hr The correct answer is: 15.5 hr