Chapter 11, Problem 94QRT

(a)

Interpretation Introduction

Interpretation:

The order of the reaction with respect to $NO$ and with respect to $O_2$ has to be determined.

Answer to Problem 94QRT

The order of the reaction with respect to $NO$ and with respect to $O_2$ is two and one respectively.

Explanation of Solution

  $2NO\left(g\right)+O_2\left(g\right)\to 2N{O}_2\left(g\right)$

Suppose the rate law of the above reaction is as follows,

  $Rate=k{\left[NO\right]}^i{\left[O_2\right]}^j$

Where, $k,iandj$ are unknown.  The values of $iandj$ can be calculated by taking given below assumptions.

Carefully examining the table, it can be said that $\left[NO\right]$ is constant in experiment number $1and3.$  Then, the rate law for experiment number $1and3$ can be written as follows,

  $\begin{array}{l}{\left(Rate\right)}_1=k{\left[NO\right]}^i{\left[O_2\right]}^j\\ \\ \Rightarrow 2.5\times {10}^{-5}mol{L}^{-1}{s}^{-1}=k{\left(0.010mol{L}^{-1}\right)}^i{\left(0.010mol{L}^{-1}\right)}^j\\ \\ {\left(Rate\right)}_3=k{\left[NO\right]}^i{\left[H_2\right]}^j\\ \\ \Rightarrow 5.0\times {10}^{-5}mol{L}^{-1}{s}^{-1}=k{\left(0.010mol{L}^{-1}\right)}^i{\left(0.020mol{L}^{-1}\right)}^j\end{array}$

Now, on dividing both the equations, the value of j can be found out.

  $\begin{array}{c}\frac{2.5\times {10}^{-5}mol{L}^{-1}{s}^{-1}}{5.0\times {10}^{-5}mol{L}^{-1}{s}^{-1}}=\frac{\cancel{k{\left(0.010mol/L\right)}^i}{\left(0.010mol{L}^{-1}\right)}^j}{\cancel{k{\left(0.010mol/L\right)}^i}{\left(0.020mol{L}^{-1}\right)}^j}\\ \\ {\left(\frac{1}{2}\right)}^1={\left(\frac{1}{2}\right)}^j\\ \\ \Rightarrow j=1.\end{array}$

Similarly, carefully examining the table, it can be said that $\left[O_2\right]$ is constant in experiment number $1and2.$  Then, the rate law for experiment number $1and2$ can be written as follows,

  $\begin{array}{l}{\left(Rate\right)}_1=k{\left[NO\right]}^i{\left[O_2\right]}^j\\ \\ \Rightarrow 2.5\times {10}^{-5}mol{L}^{-1}{s}^{-1}=k{\left(0.010mol{L}^{-1}\right)}^i{\left(0.010mol{L}^{-1}\right)}^j\\ \\ {\left(Rate\right)}_2=k{\left[NO\right]}^i{\left[H_2\right]}^j\\ \\ \Rightarrow 1.0\times {10}^{-4}mol{L}^{-1}{s}^{-1}=k{\left(0.020mol{L}^{-1}\right)}^i{\left(0.010mol{L}^{-1}\right)}^j\end{array}$

Now, on dividing both the equations, the value of i can be found out.

  $\begin{array}{c}\frac{2.5\times {10}^{-5}mol{L}^{-1}{s}^{-1}}{1.0\times {10}^{-4}mol{L}^{-1}{s}^{-1}}=\frac{{\left(0.010mol{L}^{-1}\right)}^i\cancel{k{\left(0.010mol{L}^{-1}\right)}^j}}{{\left(0.020mol{L}^{-1}\right)}^i\cancel{k{\left(0.010mol{L}^{-1}\right)}^j}}\\ \\ {\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^i\\ \\ \Rightarrow i=2.\end{array}$

Therefore, the order of the reaction with respect to $NO$ and with respect to $O_2$ is two and one respectively.

(b)

Interpretation Introduction

Interpretation:

The rate equation for the reaction has to be written.

Explanation of Solution

The order of the reaction with respect to $NO$ and with respect to $O_2$ is two and one respectively.  So the rate law can be written as given below.

  $Rate=k{\left[NO\right]}^2{\left[O_2\right]}^1$

(c)

Interpretation Introduction

Interpretation:

The rate constant $k$ has for the reaction has to be calculated.

Answer to Problem 94QRT

The average of the rate constant is $2.4\times 10^{2}Lmo{l}^{-1}{s}^{-1}.$

Explanation of Solution

The rate constant can be expressed as shown below.

  $\begin{array}{c}Rate=k{\left[NO\right]}^2{\left[O_2\right]}^1\\ \\ k=\frac{Rate}{{\left[NO\right]}^2{\left[O_2\right]}^1}\end{array}$

For first experiment the rate constant can be calculated by plugging all the data given in the table.

In first experiment, the initial concentration of $NO$ and $O_2$ are $0.010mol/Land0.010mol/L$ respectively.  The initial rate of the reaction in first experiment is $2.5\times {10}^{-5}molL{}^{-1}{s}^{-1}.$

Now, the rate constant for first experiment is given below.

  $k_1=\frac{Rate}{{\left[NO\right]}^2{\left[O_2\right]}^1}=\frac{2.5\times {10}^{-5}molL{}^{-1}{s}^{-1}}{{\left(0.010mol/L\right)}^2\left(0.010mol/L\right)}=25L^{2}mo{l}^{-2}{s}^{-1}.$

Similarly, the rate constant for rest of the experiments can be calculated.  The calculated rate constant values are given in the table below.

$\begin{array}{c}\left[NO\right]\\ \left(mol/L\right)\end{array}$	$\begin{array}{c}\left[O_2\right]\\ \left(mol/L\right)\end{array}$	Initial rate $\left(molL{}^{-1}{s}^{-1}\right)$	Rate constant $\left({\text{L}}^2{\text{mol}}^{\text{-2}}{\text{s}}^{\text{-1}}\right)$
$0.010$	$0.010$	$2.5\times {10}^{-5}$	$25$
$0.020$	$0.010$	$1.0\times {10}^{-4}$	$25$
$0.010$	$0.020$	$5.0\times {10}^{-5}$	$25$

The average of the rate constant is $25L^{2}mo{l}^{-2}{s}^{-1}.$

(d)

Interpretation Introduction

Interpretation:

The rate of the reaction has to be calculated when $\left[NO\right]$ is $0.025mol/L$ and $\left[O_2\right]$ is $0.050mol/L.$

Answer to Problem 94QRT

The rate of reaction is $7.8\times 10^{-4}mol{L}^{-1}{s}^{-1}.$

Explanation of Solution

The rate law can be expressed as shown below.

  $Rate=k{\left[NO\right]}^2{\left[O_2\right]}^1$

Given that, $\left[NO\right]$ is $0.025mol/L$ and that of $\left[O_2\right]$ is $0.050mol/L.$  The value of rate constant is $\text{25}{\text{L}}^{\text{2}}{\text{mol}}^{\text{-2}}{\text{s}}^{\text{-1}}\text{.}$  Then, by plugging all these values in the rate law, the rate of reaction can be calculated as follows,

  $\begin{array}{c}Rate=k{\left[NO\right]}^2{\left[O_2\right]}^1\\ \\ =\left(\text{25}{\text{L}}^{\text{2}}{\text{mol}}^{\text{-2}}{\text{s}}^{\text{-1}}\right){\left(0.025mol/L\right)}^2\left(0.050mol/L\right)\\ \\ =7.8\times 10^{-4}mol{L}^{-1}{s}^{-1}.\end{array}$

Therefore, the rate of reaction is $7.8\times 10^{-4}mol{L}^{-1}{s}^{-1}.$

(e)

Interpretation Introduction

Interpretation:

The rate at which $NO$ is consumed and the rate at which $N{O}_2$ is formed has to be calculated.

Concept Introduction:

Consider a reaction given below.

  $aA+bB\to cC$

Where, a, b and c are stoichiometric coefficients.  Then, the rate of disappearance of A and B with time can be written as given below.

Rate of disappearance of A$=-\frac{1}{a}\left(\frac{\Delta \left[A\right]}{\Delta t}\right)$

Rate of disappearance of B$=-\frac{1}{b}\left(\frac{\Delta \left[B\right]}{\Delta t}\right)$

Rate of appearance of C$=+\frac{1}{c}\left(\frac{\Delta \left[C\right]}{\Delta t}\right)$

Answer to Problem 94QRT

The rate at which $NO$ is consumed is $2.0\times 10^{-4}mol{L}^{-1}{s}^{-1}$ and the rate at which $N{O}_2$ is formed is $2.0\times 10^{-4}mol{L}^{-1}{s}^{-1}.$

Explanation of Solution

The reaction is given below.

  $2NO\left(g\right)+O_2\left(g\right)\to 2N{O}_2\left(g\right)$

The stoichiometric relationship is $2NO:1O_2:2N{O}_2.$

  $\begin{array}{l}-\frac{1}{2}\left(\frac{\Delta \left[NO\right]}{\Delta t}\right)=-\frac{1}{1}\left(\frac{\Delta \left[O_2\right]}{\Delta t}\right)=+\frac{1}{2}\left(\frac{\Delta \left[N{O}_2\right]}{\Delta t}\right)\\ \\ -\frac{\Delta \left[NO\right]}{\Delta t}=2\left(-\frac{\Delta \left[O_2\right]}{\Delta t}\right)=2\times \left(1.0\times {10}^{-4}mol{L}^{-1}{s}^{-1}\right)=2.0\times 10^{-4}mol{L}^{-1}{s}^{-1}\\ \\ \frac{\Delta \left[N{O}_2\right]}{\Delta t}=2\left(-\frac{\Delta \left[O_2\right]}{\Delta t}\right)=2\times \left(1.0\times {10}^{-4}mol{L}^{-1}{s}^{-1}\right)=2.0\times 10^{-4}mol{L}^{-1}{s}^{-1}\end{array}$

Therefore, the rate at which $NO$ is consumed is $2.0\times 10^{-4}mol{L}^{-1}{s}^{-1}$ and the rate at which $N{O}_2$ is formed is $2.0\times 10^{-4}mol{L}^{-1}{s}^{-1}.$