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Chapter 14: Problem 29

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} .\) The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5} .\) At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?\) (c) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M} ?\) (d) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is halved to \(0.0120 \mathrm{M} ?\)

Short Answer

Expert verified
(a) The rate law for the reaction is: Rate = k[N2O5] (b) The rate of reaction when [N2O5] = 0.0240 M is \(1.156 \times 10^{-4} \mathrm{M/s}\). (c) When the concentration of N2O5 is doubled to 0.0480 M, the rate also doubles to \(2.3136 \times 10^{-4} \mathrm{M/s}\). (d) When the concentration of N2O5 is halved to 0.0120 M, the rate halves to \(5.784 \times 10^{-5} \mathrm{M/s}\).

Step by step solution

01

Write the rate law for the reaction

Since we know that the reaction is first-order in N2O5, we can write the rate law as follows: Rate = k[N2O5]
02

Calculate the rate of the reaction when [N2O5] = 0.0240 M

We are given the rate constant, k, as \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). Using the rate law from Step 1 and the given concentration of [N2O5] = 0.0240 M, we can calculate the rate: Rate = (4.82 x 10^-3 s^-1)(0.0240 M) = \(1.156 \times 10^{-4} \mathrm{M/s}\)
03

Determine the effect of doubling the concentration of N2O5 on the rate

When the concentration of N2O5 is doubled to 0.0480 M, we can use the same rate law to calculate the new rate: New Rate = (4.82 x 10^-3 s^-1)(0.0480 M) = \(2.3136 \times 10^{-4} \mathrm{M/s}\) Notice that the new rate is exactly double the original rate (Step 2). This shows that when the concentration of a first-order reactant is doubled, the rate of the reaction also doubles.
04

Determine the effect of halving the concentration of N2O5 on the rate

When the concentration of N2O5 is halved to 0.0120 M, we can use the same rate law to calculate the new rate: New Rate = (4.82 x 10^-3 s^-1)(0.0120 M) = \(5.784 \times 10^{-5} \mathrm{M/s}\) Notice that the new rate is exactly half the original rate (Step 2). This shows that when the concentration of a first-order reactant is halved, the rate of the reaction also halves.

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Most popular questions from this chapter

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B},\) what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?

What are the differences between an intermediate and a transition state?

The following data were measured for the reaction $\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{F}_{3} \mathrm{BNH}_{3}(g):$ $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{BF}_{3}\right](M)} & {\left[\mathrm{NH}_{3}\right](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 1 & 0.250 & 0.250 & 0.2130 \\ 2 & 0.250 & 0.125 & 0.1065 \\ 3 & 0.200 & 0.100 & 0.0682 \\ 4 & 0.350 & 0.100 & 0.1193 \\ 5 & 0.175 & 0.100 & 0.0596 \\ \hline \end{array} $$ (a) What is the rate law for the reaction? (b) What is the overall order of the reaction? (c) Calculate the rate constant with proper units? (d) What is the rate when \(\left[\mathrm{BF}_{3}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{NH}_{3}\right]=0.500 \mathrm{M} ?\)

When chemists are performing kinetics experiments, the general rule of thumb is to allow the reaction to proceed for 4 half-lives. (a) Explain how you would be able to tell that the reaction has proceeded for 4 half-lives. (b) Let us suppose a reaction \(\mathrm{A} \rightarrow \mathrm{B}\) takes 6 days to proceed for 4 half-lives and is first order in A. However, when your lab partner performs this reaction for the first time, he does not realize how long it takes, and he stops taking kinetic data, monitoring the loss of A, after only 2 hours. Your lab partner concludes the reaction is zero order in A based on the data. Sketch a graph of [A] versus time to convince your lab partner the two of you need to be in the lab for a few days to obtain the proper rate law for the reaction.

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21{ }^{\circ} \mathrm{C}\). (a) Write out the balanced equation for the reaction catalyzed by urease. (b) Assuming the collision factor is the same for both situations, estimate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.
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