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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 鈮� 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]鈧�, is increased by a factor of 2.5. Let's denote the initial rate as r鈧� and the new rate as r鈧�. We have: \(r_1 = k[A_{1}]^{x}\) \(r_2 = k[A_{2}]^{x}\) Where r鈧� = initial rate, r鈧� = new rate, A鈧� = initial concentration of A and A鈧� = new concentration of A. Now, we know that r鈧� = 6.25r鈧� and A鈧� = 2.5A鈧�. We can substitute these into the equations and solve for x: \(6.25r_1 = k(2.5A_1)^x\) Divide by r鈧�: \(6.25 = (2.5)^x\) 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]鈧� is increased by a factor of 4. This means r鈧� = r鈧� or the rate remains the same. And A鈧� = 4A鈧�. 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 r鈧� = r鈧�, 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]鈧� is cut in half. So, r鈧� = (1/2)r鈧�, and A鈧� = (1/2)A鈧�. 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 r鈧�: \(\frac{1}{2} = (\frac{1}{2})^x\) 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 鈮� 2, (b) x = 0, and (c) x = 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Equation
A rate equation, also known as a rate law, describes how the rate of a chemical reaction depends on the concentration of reactants. The general form of a rate equation is given by: \[ \text{rate} = k [\text{A}]^x [\text{B}]^y \cdots \] where:
  • \(k\) is the rate constant, a proportionality factor.
  • \([\text{A}]\), \([\text{B}]\) are the concentrations of the reactants.
  • \(x\), \(y\) are the reaction orders with respect to reactants A and B.
The rate equation provides critical information about how different reactants influence the speed of the reaction. When given experimental data, you can derive the exponents like \(x\) and \(y\) by observing how changes in concentration affect the reaction rate.
Reaction Order
Reaction order refers to the exponents in the rate equation and indicates how the reaction rate is affected by the concentration of its reactants. It is the sum of all powers to which the concentration terms are raised in the rate equation. For a single-reactant reaction, it鈥檚 simply the exponent attached to the concentration of that reactant.
  • If a reaction is first-order with respect to a reactant \(\text{A}\), it means that the rate changes linearly with the change in concentration of \(\text{A}\).
  • Second-order indicates that the rate is proportional to the square of the concentration of the reactant.
  • Zero-order indicates the rate is independent of the concentration of the reactant.
In our example, different conditions provided different reaction orders for A: approximately 2 (second-order), 0 (zero-order), and 1 (first-order). Reaction order is critical in understanding how modifications in the reaction system or conditions will affect the overall reaction rate.
Concentration Effect
The concentration effect in a chemical reaction refers to how changes in the concentration of reactants affect reaction speed. This effect is directly captured in the reaction order and rate equation:
  • An increase in concentration can speed up the reaction if the reaction order is positive.
  • If the reaction order is zero, changes in concentration do not affect the rate.
  • A negative reaction order, although rare, would mean an increase in concentration actually slows the reaction down.
For instance, if changing the concentration of a reactant does not change the rate, the concentration effect is zero, as seen in Case (b). Understanding how concentration affects rate is essential, especially in industrial applications where reaction speed is crucial. These insights enable chemists to manipulate reaction conditions to optimize processes for faster or more controlled reactions.

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Most popular questions from this chapter

(a) Develop an equation for the half-life of a zero-order reaction. (b) Does the half-life of a zero-order reaction increase, decrease, or remain the same as the reaction proceeds?

(a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products?

Which of the following linear plots do you expect for a reaction \(\mathrm{A} \longrightarrow\) products if the kinetics are (a) zero order, (b) first order, or (c) second order?

(a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) Does the rate constant for a reaction generally increase or decrease with an increase in reaction temperature? (c) Which factor is most sensitive to changes in temperature-the frequency of collisions, the orientation factor, or the fraction of molecules with energy greater than the activation energy?

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 \(\left[\mathrm{OH}^{-}\right]=0.050 \mathrm{M}\)
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