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Chapter 14: Problem 92

You perform a series of experiments for the reaction $\mathrm{A} \rightarrow 2 \mathrm{~B}$ and find that the rate law has the form, rate \(=k[\mathrm{~A}]^{x} .\) Determine the value of \(x\) in each of the following cases: (a) The rate increases by a factor of \(6.25,\) when \([\mathrm{A}]_{0}\) is increased by a factor of \(2.5 .(\mathbf{b})\) There is no rate change when \([\mathrm{A}]_{0}\) is increased by a factor of \(4 .(\mathbf{c})\) The rate decreases by a factor of \(1 / 2,\) when \([\mathrm{A}]_{0}\) is cut in half.

Short Answer

Expert verified
In summary, the values of x for each case are: (a) x ≈ 2, (b) x = 0, and (c) x = 1.

Step by step solution

01

Case (a)

We are given that the rate increases by a factor of 6.25 when the initial concentration of A, [A]₀, is increased by a factor of 2.5. Let's denote the initial rate as r₁ and the new rate as r₂. We have: \(r_1 = k[A_{1}]^{x}\) \(r_2 = k[A_{2}]^{x}\) Where r₁ = initial rate, r₂ = new rate, A₁ = initial concentration of A and A₂ = new concentration of A. Now, we know that r₂ = 6.25r₁ and A₂ = 2.5A₁. We can substitute these into the equations and solve for x: \(6.25r_1 = k(2.5A_1)^x\) Divide by r₁: \(6.25 = (2.5)^x\) Now, take the logarithm on both sides: \(\log{6.25} = x\log{2.5}\) Then solve for x: \(x = \frac{\log{6.25}}{\log{2.5}} \approx 2\) So, the value of x in case (a) is approximately 2.
02

Case (b)

We are given that there is no rate change when [A]₀ is increased by a factor of 4. This means r₂ = r₁ or the rate remains the same. And A₂ = 4A₁. We can substitute these into the equations and solve for x: \(r_1 = k[A_{1}]^{x}\) \(r_1 = k[4A_{1}]^{x}\) Now, since r₁ = r₂, we can divide the second equation by the first equation: \(1 = (4)^x\) Taking the logarithm on both sides, we have: \(0 = x\log{4}\) Then solve for x: \(x = \frac{0}{\log{4}} = 0\) So, the value of x in case (b) is 0.
03

Case (c)

We are given that the rate decreases by a factor of 1/2 when [A]₀ is cut in half. So, r₂ = (1/2)r₁, and A₂ = (1/2)A₁. We can substitute these into the equations and solve for x: \(\frac{1}{2}r_1 = k(\frac{1}{2}A_1)^x\) Divide by r₁: \(\frac{1}{2} = (\frac{1}{2})^x\) Taking the logarithm on both sides: \(\log{\frac{1}{2}} = x\log{\frac{1}{2}}\) Then solve for x: \(x = \frac{\log{\frac{1}{2}}}{\log{\frac{1}{2}}} = 1\) So, the value of x in case (c) is 1. In summary, the values of x for each case are: (a) x ≈ 2, (b) x = 0, and (c) x = 1.

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Most popular questions from this chapter

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\) $3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) .\( If the concentration of \)\mathrm{C}_{2} \mathrm{H}_{4}$ is decreasing at the rate of \(0.025 \mathrm{M} / \mathrm{s}\), what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and $\mathrm{H}_{2} \mathrm{O}\( ? (b) The rate of decrease in \)\mathrm{N}_{2} \mathrm{H}_{4}$ partial pressure in a closed reaction vessel from the reaction $\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$ is \(10 \mathrm{kPa}\) per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

The reaction $2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\( \)\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ was studied with the following results: $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{CIO}_{2}\right](M)} & {\left[\mathrm{OH}^{-}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1 & 0.060 & 0.030 & 0.0248 \\ 2 & 0.020 & 0.030 & 0.00276 \\ 3 & 0.020 & 0.090 & 0.00828 \\ \hline \end{array} $$ (a) Determine the rate law for the reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{ClO}_{2}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]=0.050 \mathrm{M}\)

(a) A certain first-order reaction has a rate constant of $2.75 \times 10^{-2} \mathrm{~s}^{-1}\( at \)20^{\circ} \mathrm{C}\(. What is the value of \)k$ at \(60^{\circ} \mathrm{C}\) if $E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{b})\( Another first-order reaction also has a rate constant of \)2.75 \times 10^{-2} \mathrm{~s}^{-1}\( at \)20^{\circ} \mathrm{C}$. What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if $E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{c})$ What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

The activation energy of an uncatalyzed reaction is $95 \mathrm{~kJ} / \mathrm{mol}\(. The addition of a catalyst lowers the activation energy to \)55 \mathrm{~kJ} / \mathrm{mol}$. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C},(\mathbf{b}) 125^{\circ} \mathrm{C} ?\)

Consider the following hypothetical aqueous reaction: $\mathrm{A}(a q) \rightarrow \mathrm{B}(a q)\(. A flask is charged with \)0.065 \mathrm{~mol}$ of \(\mathrm{A}\) in a total volume of \(100.0 \mathrm{~mL}\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 10 & 20 & 30 & 40 \\ \hline \text { Moles of A } & 0.065 & 0.051 & 0.042 & 0.036 & 0.031 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that there are no molecules of \(\mathrm{B}\) at time zero and that A cleanly converts to B with no intermediates. (b) Calculate the average rate of disappearance of A for each 10 -min interval in units of \(M /\) s. (c) Between \(t=0 \mathrm{~min}\) and \(t=30 \mathrm{~min},\) what is the average rate of appearance of \(\mathrm{B}\) in units of \(\mathrm{M} / \mathrm{s}\) ? Assume that the volume of the solution is constant.
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