Chapter 11, Problem 27QAP

Hydrogen bromide is a highly reactive and corrosive gas used mainly as a catalyst for organic reactions. It is produced by reacting hydrogen and bromine gases together.

  ${\text{H}}_2\left(g\right)+{\text{Br}}_2\left(g\right)\to 2\text{HBr}\left(g\right)$The rate is followed by measuring the intensity of the orange color of the bromine gas. The following data are obtained:

(a) What is the order of the reaction with respect to hydrogen, bromine, and overall?

(b) Write the rate expression of the reaction.

(c) Calculate k for the reaction. What are the units for k?

(d) When $\left[{\text{H}}_2\right]=0.455M$and $\left[{\text{Br}}_2\right]=0.215M$, what is the rate of the reaction?

Interpretation Introduction

(a)

Interpretation:

To determine the order of reaction with respect to BF₃, NH₃ and overall for the following reaction:

${\text{H}}_{\text{2}}\text{(g)}\text{+}{\text{Br}}_{\text{2}}\text{(g)}\stackrel{}{\to }\text{2HBr(g)}$

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 27QAP

Order of given reaction:

With respect to H₂ =1

With respect to Br₂ = $\frac{1}{2}$

Overall = $\frac{\text{3}}{2}$.

Explanation of Solution

Given information:

Here the chemical reaction is:

${\text{H}}_{\text{2}}\text{(g)}\text{+}{\text{Br}}_{\text{2}}\text{(g)}\stackrel{}{\to }\text{2HBr(g)}$

Let’s assume the reaction to be ‘t’ order with respect to H₂ and ‘y’ order with respect to Br₂.

Then, rate law for experiment 1 in above reaction will be;

$\begin{array}{l}\text{rate}\text{=}{\text{k[H}}_{\text{2}}{\text{]}}^{\text{t}}{{\text{[Br}}_{\text{2}}}^{\text{]}}\\ \text{4}{\text{.74×10}}^{\text{-3}}\text{=}\text{k[0}{\text{.100]}}^{\text{t}}{\text{[0}}^{\text{.100]}}\mathrm{......}\text{(1)}\end{array}$

And, rate law for experiment 3 in above reaction will be;

$\begin{array}{l}\text{rate}\text{=}{\text{k[H}}_{\text{2}}{\text{]}}^{\text{t}}{{\text{[Br}}_{\text{2}}}^{\text{]}}\\ \text{6}{\text{.71×10}}^{\text{-3}}\text{=}\text{k[0}{\text{.100]}}^{\text{t}}{\text{[0}}^{\text{.200]}}\mathrm{......}\text{(2)}\end{array}$

Divide (1) by (2) to get value of ‘t’.

$\begin{array}{l}\frac{\text{4}\text{.74}\times {\text{10}}^{\text{-3}}}{\text{6}\text{.71}\times {\text{10}}^{\text{-3}}}\text{=}{\left(\frac{\text{0}\text{.100}}{\text{0}\text{.200}}\right)}^y\\ \Rightarrow 0.7\text{=}{\left(0.5\right)}^y\\ \text{taking}\text{log}\text{both}\text{sides,}\\ \text{log(0}\text{.7)}\text{=}\text{y×log(0}\text{.5)}\left({\text{log(a)}}^{\text{t}}\text{=}\text{t×log(a)}\right)\\ y=\frac{(-0.15)}{(-0.30)}=\frac{1}{2}\end{array}$

Thus, order with respect to Br₂ is $\frac{1}{2}$

Now, rate law for experiment 4 in above reaction will be;

$\begin{array}{l}\text{rate}\text{=}{\text{k[H}}_{\text{2}}{\text{]}}^{\text{t}}{{\text{[Br}}_{\text{2}}}^{\text{]}}\\ \text{1}{\text{.68×10}}^{\text{-2}}\text{=}\text{k[0}{\text{.250]}}^{\text{t}}{\text{[0}}^{\text{.200]}}\mathrm{......}\text{(3)}\end{array}$

Divide (2) by (3) to get value of ‘y’.

$\begin{array}{l}\frac{\text{6}\text{.71}\times {\text{10}}^{\text{-3}}}{1.68\times {\text{10}}^{\text{-2}}}\text{=}{\left(\frac{\text{0}\text{.100}}{\text{0}\text{.250}}\right)}^{\text{t}}\\ \Rightarrow 0.4\text{=}{\left(\text{0}\text{.4}\right)}^{\text{t}}\\ \Rightarrow \text{t=}\text{1}\end{array}$

Thus, order with respect to H₂ is 1.

And the order of reaction will be:

$\Rightarrow \text{ t + y = }\frac{\text{1}}{2}\text{+1 =}\frac{\text{3}}{2}$

Thus, overall order of reaction is 2.

Interpretation Introduction

(b)

Interpretation:

To write the rate expression for the given reaction.

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 27QAP

Rate law expression for above reaction will be;

$\text{rate}\text{=}{\text{k[H}}_{\text{2}}{\text{][Br}}_{\text{2}}{\text{]}}^{\frac{1}{2}}$

Explanation of Solution

Here the chemical reaction is:

${\text{H}}_{\text{2}}\text{(g)}\text{+}{\text{Br}}_{\text{2}}\text{(g)}\stackrel{}{\to }\text{2HBr(g)}$

Order of reaction with respect to H₂ = 2

Order of reaction with respect to Br₂ = $\frac{1}{2}$

Let the rate constant be ‘k’.

Then, rate law expression for above reaction will be;

$\begin{array}{l}\text{rate}\text{=}{\text{k[H}}_{\text{2}}{\text{][Br}}_{\text{2}}{\text{]}}^{\frac{1}{2}}\\ \end{array}$

Interpretation Introduction

(c)

Interpretation:

To determine the rate constant and its unit for the given reaction.

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 27QAP

Rate constant is $0.15{\text{L}}^{\frac{\text{1}}{2}}{\text{/mol}}^{\frac{\text{1}}{2}}\text{.s}$

Unit of rate constant is ${\text{L}}^{\frac{\text{1}}{2}}{\text{.mol}}^{-\frac{\text{1}}{2}}{\text{.s}}^{\text{-1}}$

Explanation of Solution

Here the chemical reaction is:

${\text{H}}_{\text{2}}\text{(g)}\text{+}{\text{Br}}_{\text{2}}\text{(g)}\stackrel{}{\to }\text{2HBr(g)}$

Writing rate law for experiment 1 in above reaction will be;

$\begin{array}{l}\text{rate}\text{=}{\text{k[H}}_{\text{2}}{\text{][Br}}_{\text{2}}{\text{]}}^{\frac{\text{1}}{\text{2}}}\\ \text{4}{\text{.74×10}}^{\text{-3}}\left(\frac{\text{mol}}{\text{L}\text{.s}}\right)\text{=}\text{k}\left(\text{U}\right)\text{×0}\text{.100}\left(\frac{\text{mol}}{\text{L}}\right)\text{×}{\left[\text{0}\text{.100}\left(\frac{\text{mol}}{\text{L}}\right)\right]}^{\frac{\text{1}}{\text{2}}}\\ \text{where}\text{U}\text{represent}\text{unit}\text{of}\text{rate}\text{constant}\text{.}\\ \Rightarrow \text{k}\left(\text{U}\right)\text{=}\frac{\text{4}{\text{.74×10}}^{\text{-3}}\left(\frac{\text{mol}}{\text{L}\text{.s}}\right)}{\text{0}\text{.100}\left(\frac{\text{mol}}{\text{L}}\right)\times 0.316{\left(\frac{\text{mol}}{\text{L}}\right)}^{\frac{1}{2}}}=0.15{\text{L}}^{\frac{\text{1}}{2}}{\text{/mol}}^{\frac{\text{1}}{2}}\text{.s}\\ \end{array}$

Hence, the rate constant is $0.15{\text{L}}^{\frac{\text{1}}{2}}{\text{/mol}}^{\frac{\text{1}}{2}}\text{.s}$

And unit of rate constant is ${\text{L}}^{\frac{\text{1}}{2}}{\text{.mol}}^{-\frac{\text{1}}{2}}{\text{.s}}^{\text{-1}}$

Interpretation Introduction

(d)

Interpretation:

To determine the rate of reaction at given concentration of reactants.

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 27QAP

Rate of reaction for given reaction at given conditions is $0.029\text{mol/L}\text{.s}$

Explanation of Solution

Here the chemical reaction is:

${\text{H}}_{\text{2}}\text{(g)}\text{+}{\text{Br}}_{\text{2}}\text{(g)}\stackrel{}{\to }\text{2HBr(g)}$

Rate law expression for above reaction:

$\text{rate}\text{=}{\text{k[H}}_{\text{2}}{\text{][Br}}_{\text{2}}{\text{]}}^{\frac{\text{1}}{\text{2}}}$

Here we have:

[H₂ ]=0.421 M

[Br₂ ] = 0.215 M

Rate constant = $0.15{\text{L}}^{\frac{\text{1}}{2}}{\text{/mol}}^{\frac{\text{1}}{2}}\text{.s}$

Plugging values in rate law as:

$\begin{array}{l}\text{rate}\text{=}0.15\times (\text{0}\text{.421)}\times {(}^0\\ \text{rate}\text{=}0.029\text{mol/L}\text{.s}\end{array}$

Hence, the rate of reaction is $0.029\text{mol/L}\text{.s}$