Chapter 11, Problem 25QAP

When boron trifluoride reacts with ammonia, the following reaction occurs:

${\text{BF}}_3\left(g\right)+{\text{NH}}_3\left(g\right)\to {\text{BF}}_3{\text{NH}}_3\left(g\right)$The following data are obtained at a particular temperature:

(a) What is the order of the reaction with respect to BF₃, NH₃, and overall?

(b) Write the rate expression for the reaction.

(c) Calculate k for the reaction.

(d) When $\left[{\text{BF}}_3\right]=0.533M$ and ${\text{NH}}_3=0.300M$, what is the rate of the reaction at the temperature of the experiment?

Interpretation Introduction

(a)

Interpretation:

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

${\text{BF}}_{\text{3}}\text{(g)}\text{+}{\text{NH}}_{\text{3}}\text{(g)}\stackrel{}{\to }{\text{BF}}_{\text{3}}{\text{NH}}_{\text{3}}\text{(g)}$

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 25QAP

Order of given reaction

With respect to BF₃ =1

With respect to NF₃ =1

Overall = 2.

Explanation of Solution

Given information:

Here the chemical reaction is:

${\text{BF}}_{\text{3}}\text{(g)}\text{+}{\text{NH}}_{\text{3}}\text{(g)}\stackrel{}{\to }{\text{BF}}_{\text{3}}{\text{NH}}_{\text{3}}\text{(g)}$

Let’s assume the reaction to be ‘t’ order with respect to BF₃ and ‘y’ order with respect to NH₃.

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

$\begin{array}{l}\text{rate}\text{=}{\text{k[BF}}_{\text{3}}{\text{]}}^{\text{t}}{{\text{[NH}}_{\text{3}}}^{\text{]}}\\ \text{0}\text{.0341}\text{=}\text{k[0}{\text{.100]}}^{\text{t}}{\text{[0}}^{\text{.100]}}\mathrm{......}(1)\end{array}$

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

$\begin{array}{l}\text{rate}\text{=}{\text{k[BF}}_{\text{3}}{\text{]}}^{\text{t}}{{\text{[NH}}_{\text{3}}}^{\text{]}}\\ \text{0}\text{.102}\text{=}\text{k[0}{\text{.300]}}^{\text{t}}{\text{[0}}^{\text{.100]}}\mathrm{......}(2)\end{array}$

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

$\begin{array}{l}\frac{\text{0}\text{.0341}}{\text{0}\text{.102}}\text{=}{\left(\frac{\text{0}\text{.100}}{\text{0}\text{.300}}\right)}^{\text{t}}\\ \Rightarrow \text{0}\text{.334}\text{=}{\left(\frac{\text{1}}{\text{3}}\right)}^{\text{t}}\\ \Rightarrow \text{taknig}\text{log}\text{both}\text{side,}\text{we}\text{get:}\\ \text{log(0}\text{.334)}\text{=log}{\left(\frac{\text{1}}{\text{3}}\right)}^{\text{t}}\\ \text{As}\text{we}\text{know:}\\ \text{l}\text{og}{\left(\text{a}\right)}^{\text{t}}\text{=}\text{t×log(a)}\\ \Rightarrow \text{-0}\text{.476=t×}\text{log}\left(\frac{\text{1}}{\text{3}}\right)\\ \Rightarrow \text{-0}\text{.476}\text{=}\text{t×[log(1)}\text{-}\text{log(3)]}\\ \Rightarrow \text{-0}\text{.476}\text{=}\text{t×[0-0}\text{.476]}\\ \Rightarrow \text{0}\text{.476}\text{=}\text{t}\text{×0}\text{.476}\\ \Rightarrow \text{t=1}\end{array}$

Now writing rate law for experiment 2 in above reaction will be;

$\begin{array}{l}\text{rate}\text{=}{\text{k[BF}}_{\text{3}}{\text{]}}^{\text{t}}{{\text{[NH}}_{\text{3}}}^{\text{]}}\\ \text{0}\text{.159}\text{=}\text{k[0}{\text{.200]}}^{\text{t}}{\text{[0}}^{\text{.233]}}\mathrm{......}(3)\end{array}$

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

$\begin{array}{l}\text{rate}\text{=}{\text{k[BF}}_{\text{3}}{\text{]}}^{\text{t}}{{\text{[NH}}_{\text{3}}}^{\text{]}}\\ \text{0}\text{.0512}\text{=}\text{k[0}{\text{.200]}}^{\text{t}}{\text{[0}}^{\text{.0750]}}\mathrm{......}(4)\end{array}$

Divide (3) by (4) to get value of ‘y’.

$\begin{array}{l}\text{3}\text{.10}\text{=}{\left(\frac{\text{0}\text{.233}}{\text{0}\text{.0750}}\right)}^{\text{y}}\\ \Rightarrow \text{3}\text{.10}\text{=}{\left(\text{3}\text{.10}\right)}^{\text{y}}\\ \Rightarrow \text{y}\text{=}\text{1}\end{array}$

Thus, order with respect to NH₃ is 1.

And the order of reaction will be:

$\Rightarrow t+y=1+1=2$

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 25QAP

Rate law expression for the given reaction will be;

$\text{rate}\text{=}{\text{k[BF}}_{\text{3}}{\text{][NH}}_{\text{3}}\text{]}$

Explanation of Solution

Here the chemical reaction is:

${\text{BF}}_{\text{3}}\text{(g)}\text{+}{\text{NH}}_{\text{3}}\text{(g)}\stackrel{}{\to }{\text{BF}}_{\text{3}}{\text{NH}}_{\text{3}}\text{(g)}$

Order of reaction with respect to BF₃ = 1

Order of reaction with respect to NH₃ = 1

Let the rate constant be ‘k’.

Then, rate law expression for above reaction will be;

$\text{rate}\text{=}{\text{k[BF}}_{\text{3}}{\text{][NH}}_{\text{3}}\text{]}$

Interpretation Introduction

(c)

Interpretation:

To determine the rate constant 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 25QAP

Rate constant for given reaction is $3.41\text{L/mol}\text{.s}$

Explanation of Solution

Here the chemical reaction is:

${\text{BF}}_{\text{3}}\text{(g)}\text{+}{\text{NH}}_{\text{3}}\text{(g)}\stackrel{}{\to }{\text{BF}}_{\text{3}}{\text{NH}}_{\text{3}}\text{(g)}$

Rate law expression for above reaction:

$\text{rate}\text{=}{\text{k[BF}}_{\text{3}}{\text{][NH}}_{\text{3}}\text{]}$

Plugging values in rate law expression for experiment 1 as:

$0.0341\text{=}\text{k[0}\text{.100][0}\text{.100]}$

$\begin{array}{l}\Rightarrow \text{k}\text{=}\frac{\text{0}\text{.0341}}{\text{0}\text{.100}\times \text{0}\text{.100}}\\ \Rightarrow \text{k}\text{=}3.41\text{L/mol}\text{.s}\end{array}$

Hence, the rate constant is $3.41\text{L/mol}\text{.s}$

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 25QAP

Rate of reaction for given reaction at given conditions is $\text{0}\text{.5453}\text{mol/L}\text{.s}$

Explanation of Solution

Here the chemical reaction is:

${\text{BF}}_{\text{3}}\text{(g)}\text{+}{\text{NH}}_{\text{3}}\text{(g)}\stackrel{}{\to }{\text{BF}}_{\text{3}}{\text{NH}}_{\text{3}}\text{(g)}$

Rate law expression for above reaction:

$\text{rate}\text{=}{\text{k[BF}}_{\text{3}}{\text{][NH}}_{\text{3}}\text{]}$

Here we have:

[BF₃ ]= 0.533 M

[NH₃ ] = 0.300 M

Rate constant = 3.41 L/mol.s

Plugging values in rate law as:

$\begin{array}{l}\text{rate=}\text{3}\text{.41[0}\text{.533][0}\text{.300]}\\ \text{rate}\text{=}\text{0}\text{.5453}\text{mol/L}\text{.s}\end{array}$

Hence, the rate of reaction is $\text{0}\text{.5453}\text{mol/L}\text{.s}$