Chemical Kinetics — Practice

Question 1

For the reaction N2 + 3H2 gives 2NH3, ammonia is formed at 0.06 mol/L/s. Find the rate of the reaction and the rate at which hydrogen is consumed.

Accepted Answer

Divide by each species' coefficient to get the single rate of reaction. Answer: Rate = 0.03 mol/L/s; hydrogen is consumed at 0.09 mol/L/s.

Question 2

A reaction has rate constant 3.2 x 10^-4 L mol-1 s-1. What is its order, and why?

Accepted Answer

Match the units against the general expression for the units of k. Answer: Second order, because L/mol/s is (mol/L)^-1 per second, which needs n = 2.

Question 3

A first order reaction has a half-life of 20 minutes. What fraction of the reactant remains after one hour?

Accepted Answer

One hour is a whole number of half-lives. Answer: Three half-lives, so one eighth, or 12.5%, remains.

Question 4

A first order reaction is 20% complete in 10 minutes. Calculate the rate constant and the time for 75% completion.

Accepted Answer

Use the amount remaining in the logarithm, not the amount consumed. Answer: k = 2.23 x 10^-2 per min; t = 62.1 min.

Question 5

For a reaction A + B gives products, doubling [A] doubles the rate and doubling [B] leaves it unchanged. Write the rate law and state the overall order.

Accepted Answer

Read each order off the pair of experiments that holds the other concentration fixed. Answer: Rate = k[A]; overall order 1.

Question 6

The rate constant of a reaction is 1.5 x 10^-3 per second at 300 K and 6.0 x 10^-3 per second at 320 K. Calculate the activation energy.

Accepted Answer

Use the two-temperature form, in which the pre-exponential factor cancels. Answer: Ea = 55.3 kJ/mol.

Question 7

A zero order reaction has k = 0.02 mol/L/s and an initial concentration of 0.5 mol/L. Find the half-life and the time for the reaction to go to completion.

Accepted Answer

Zero order concentration falls linearly, so both times follow from the straight line. Answer: t(1/2) = 12.5 s; complete in 25 s.

Question 8

A reaction has an activation energy of 209 kJ/mol and a rate constant of 1.6 x 10^-5 per second at 600 K. At what temperature will the rate constant be 100 times larger?

Accepted Answer

Use the two-temperature form with k2/k1 = 100 and solve for the reciprocal temperature. Answer: About 730 K.

Question 9

For the reaction 2NO + O2 gives 2NO2, doubling [NO] at fixed [O2] multiplies the rate by four, and doubling [O2] at fixed [NO] doubles it. Write the rate law, give the overall order and the units of k, and comment on the molecularity.

Accepted Answer

Deduce each exponent separately, then compare the result with the balanced equation. Answer: Rate = k[NO]^2[O2]; third order; units mol-2 L2 s-1. The exponents happen to match the coefficients here, but that is a coincidence of this reaction, not a rule.

Question 10

For the gas-phase first order decomposition X(g) gives 2Y(g) at constant volume, the initial pressure of X is 0.4 atm and the total pressure after 300 s is 0.55 atm. Calculate the rate constant.

Accepted Answer

Express the remaining pressure of X in terms of the total pressure before using the integrated law. Answer: p(X) = 0.8 - P = 0.25 atm; k = 1.57 x 10^-3 per second.

Question 11

For a reaction with a rate law Rate = k[A]^2[B]^-1, describe how the rate changes when [A] is doubled and [B] is tripled simultaneously.

Accepted Answer

A negative order means the species slows the reaction down. Answer: The rate changes by a factor of 4/3, so it increases by about 33%.

Practice — Chemical Kinetics

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