Chapter 14, Problem 5QP

Write the reaction rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of products:

(a)

${\text{H}}_{\text{2}}\left(g\right)+{\text{l}}_{\text{2}}\left(g\right)\to \text{2Hl}\left(g\right)$

(b)

${\text{5Br}}^{\text{-}}\left(aq\right)+{\text{BrO}}_3^{-}\left(aq\right)+6{\text{H}}^{\text{+}}\left(aq\right)\to {\text{3Br}}_{\text{2}}\left(aq\right)+{\text{3H}}_{\text{2}}\text{O}\left(l\right)$

Interpretation Introduction

Interpretation: The rate expression of the reactions, in terms of disappearance for the reactants and appearance of the product, is to be written.

Concept introduction: The rate of the consumption of reactants is equal to the rate of the formation of products. The disappearance of reactants is represented by a negative sign, while the appearance of products is represented by a positive sign, in the rate law expression.

Answer to Problem 5QP

Solution:

a)

$\text{Rate}=-\frac{\Delta \left[{\text{H}}_2\right]}{\Delta t}=-\frac{\Delta \left[{\text{I}}_2\right]}{\Delta t}=\frac{1}{2}\frac{\Delta \left[\text{HI}\right]}{\Delta t}$

b)

$\text{Rate}=-\frac{1}{5}\frac{\Delta \left[{\text{Br}}^{-}\right]}{\Delta t}=-\frac{\Delta \left[{\text{BrO}}_3^{-}\right]}{\Delta t}=-\frac{1}{6}\frac{\Delta \left[{\text{H}}^{+}\right]}{\Delta t}=\frac{1}{3}\frac{\Delta \left[{\text{Br}}_2\right]}{\Delta t}=\frac{1}{3}\frac{\Delta \left[{\text{H}}_{\text{2}}\text{O}\right]}{\Delta t}$

Explanation of Solution

(a)

Given information: Equation: $H_2\left(g\right)+I_2\left(g\right)\to 2HI\left(g\right)$.

For a general reaction given as follows:

$a\text{A}+b\text{B}\to c\text{C}+d\text{D}$

The rate of reaction, in terms of disappearance of reactants and appearance of products, is written as follows:

$\text{Rate}=\frac{1}{a}\frac{\Delta \left[\text{A}\right]}{\Delta t}=\frac{1}{b}\frac{\Delta \left[\text{B}\right]}{\Delta t}=-\frac{1}{c}\frac{\Delta \left[\text{C}\right]}{\Delta t}=-\frac{1}{d}\frac{\Delta \left[\text{D}\right]}{\Delta t}$

The rate expression, in terms of disappearance of hydrogen and iodine gas, and appearance of hydrogen iodide gas, is written as follows:

$\text{Rate}=-\frac{\Delta \left[{\text{H}}_2\right]}{\Delta t}=-\frac{\Delta \left[{\text{I}}_2\right]}{\Delta t}=\frac{1}{2}\frac{\Delta \left[\text{HI}\right]}{\Delta t}$

(b)

Given information: Equation: $5B{r}^{-}\left(aq\right)+BrO\left(aq\right)+6H^{+}\left(aq\right)\to 3B{r}_2\left(aq\right)+3H_{2}O\left(l\right)$

For a general reaction given as follows:

$a\text{A}+b\text{B}\to c\text{C}+d\text{D}$

The rate of reaction, in terms of disappearance of reactants and appearance of products, is written as follows:

$\text{Rate}=\frac{1}{a}\frac{\Delta \left[\text{A}\right]}{\Delta t}=\frac{1}{b}\frac{\Delta \left[\text{B}\right]}{\Delta t}=-\frac{1}{c}\frac{\Delta \left[\text{C}\right]}{\Delta t}=-\frac{1}{d}\frac{\Delta \left[\text{D}\right]}{\Delta t}$

The rate expression, in terms of disappearance of ${\text{Br}}^{-},{\text{BrO}}_3^{-}$, and hydrogen ions, and appearance of ${\text{Br}}_2$ and water, is written as follows:

$\text{Rate}=-\frac{1}{5}\frac{\Delta \left[{\text{Br}}^{-}\right]}{\Delta t}=-\frac{\Delta \left[{\text{BrO}}_3^{-}\right]}{\Delta t}=-\frac{1}{6}\frac{\Delta \left[{\text{H}}^{+}\right]}{\Delta t}=\frac{1}{3}\frac{\Delta \left[{\text{Br}}_2\right]}{\Delta t}=\frac{1}{3}\frac{\Delta \left[{\text{H}}_{\text{2}}\text{O}\right]}{\Delta t}$