Consider the reaction $$ \mathrm{X}+\mathrm{Y} \longrightarrow \mathrm{Z} $$ From the following data, obtained at \(360 \mathrm{~K}\), (a) determine the order of the reaction, and (b) determine the initial rate of disappearance of \(\mathrm{X}\) when the concentration of \(\mathrm{X}\) is \(0.30 \mathrm{M}\) and that of Y is \(0.40 \mathrm{M}\). $$ \begin{array}{ccc} \hline \text { Initial Rate of } & & \\ \text { Disappearance of } \mathbf{X}(\boldsymbol{M} / \mathbf{s}) & {[\mathrm{X}](M)} & {[\mathrm{Y}](M)} \\ \hline 0.053 & 0.10 & 0.50 \\ 0.127 & 0.20 & 0.30 \\ 1.02 & 0.40 & 0.60 \\ 0.254 & 0.20 & 0.60 \\ 0.509 & 0.40 & 0.30 \\ \hline \end{array} $$

Short Answer

Expert verified
The reaction is first order with respect to X and zeroth order with respect to Y. The initial rate of disappearance of reactant X when its concentration is 0.30 M and Y is 0.40 M is 0.159 M/s.

Step by step solution

01

Determine the reaction order with respect to X

First, we need to compare two reactions where only the concentration of X changes, and the concentration of Y stays constant. Reactions 2 and 4 fulfill these conditions. The rate increases by a factor of \( r = 0.254 \, M/s ÷ 0.127 \, M/s = 2 \) when the concentration of X doubles (from 0.20 M to 0.40 M). If the reaction was first order with respect to X, the rate would also double. Since it does, the reaction is first order with respect to X.
02

Determine the reaction order with respect to Y

Next, we need to compare two reactions where only the concentration of Y changes, and the concentration of X stays constant. Reactions 1 and 2 fulfill these conditions. The rate increases by a factor of \( r = 0.127 \, M/s ÷ 0.053 \, M/s = 2.396 \) when the concentration of Y decreases by a factor of \( f = 0.30 \, M ÷ 0.50 \, M = 0.6 \). Since the rate of the reaction does not decline with Y by the same factor (is not proportional), the reaction is zeroth-order with respect to Y.
03

Determine the reaction rate constant

For a first-order dependence on [X] and zeroth-order dependence on [Y], the rate law is \( Rate = k[X] \). We can use any of the provided reactions to solve for the rate constant k. Using reaction 1: \( k = Rate ÷ [X] = 0.053\, M/s ÷ 0.10\, M = 0.53\, s^{-1} \)
04

Calculate the initial rate of disappearance of X for given concentrations

Now, with the rate constant and order of the reaction known, we can calculate the initial rate of disappearance of X when [X] is 0.30 M and [Y] is 0.40 M: \( Rate = k[X] = 0.53\, s^{-1} × 0.30\, M = 0.159\, M/s \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The decomposition of dinitrogen pentoxide has been studied in carbon tetrachloride solvent \(\left(\mathrm{CCl}_{4}\right)\) at a certain temperature: $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} $$ $$ \begin{array}{cc} \hline\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 0.92 & 0.95 \times 10^{-5} \\ 1.23 & 1.20 \times 10^{-5} \\ 1.79 & 1.93 \times 10^{-5} \\ 2.00 & 2.10 \times 10^{-5} \\ 2.21 & 2.26 \times 10^{-5} \\ \hline \end{array} $$ Determine graphically the rate law for the reaction and calculate the rate constant.

The rate constant for the second-order reaction $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is \(0.54 / M \cdot \mathrm{s}\) at \(300^{\circ} \mathrm{C}\). How long (in seconds) would it take for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(0.62 M\) to \(0.28 M ?\)

The equation for the combustion of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) is $$ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Explain why it is unlikely that this equation also represents the elementary step for the reaction.

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ fructose \(\quad\) glucose This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) even though water is a reactant. $$ \begin{array}{cc} \hline \text { Time (min) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]} \\ \hline 0 & 0.500 \\ 60.0 & 0.400 \\ 96.4 & 0.350 \\ 157.5 & 0.280 \\ \hline \end{array} $$

For a first-order reaction, how long will it take for the concentration of reactant to fall to one-eighth its original value? Express your answer in terms of the half-life \(\left(t_{1}\right)\) and in terms of the rate constant \(k\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free