Chapter 11, Problem 100QRT

(a)

Interpretation Introduction

Interpretation:

Whether the given reaction is zeroth, first-, or second-order by using the units of the rate constant has to be determined.

Concept Introduction:

  $\begin{array}{cc}\text{order}\text{of}\text{reaction}& \text{unit}\text{of}\text{rate}\text{constant}\\ 0& mol{L}^{-1}tim{e}^{-1}\\ 1& tim{e}^{-1}\\ 2& mo{l}^{-1}Ltim{e}^{-1}\end{array}$

Explanation of Solution

The rate constant of the reaction is $0.68s^{-1}.$ So the given reaction is first order because the unit of its rate constant is $time{}^{-1}.$

(b)

Interpretation Introduction

Interpretation:

The mass of azomethane remains after $5.0s$ has to be calculated.

Concept Introduction:

The integrated rate equation for first order reaction is given below.

  $\lambda t=\ln \frac{{\left[A\right]}_o}{{\left[A\right]}_t}$

Where,

    $\lambda$ is the rate constant/ decay constant

    $t$ is the time taken to change the concentration from ${\left[A\right]}_{o}to{\left[A\right]}_t$

    ${\left[A\right]}_o$ is the initial concentration of $A$

    ${\left[A\right]}_t$ is the final concentration $A$

Half-life of a first order reaction can be calculated using following equation.

    $t_{\frac{1}{2}}=\frac{0.693}{\lambda }$

Answer to Problem 100QRT

The mass of azomethane remains after $5.0s$ is $0.65g.$

Explanation of Solution

Given,

    $\lambda =0.68s^{-1}.$

    $t=5.0s$

    ${\left[A\right]}_o=\frac{\frac{2g}{58.08g/mol}}{2L}=0.17M$

The mass of azomethane remains after $5.0s$ can be calculated as follows,

    $\begin{array}{c}\lambda t=\ln \frac{{\left[A\right]}_o}{{\left[A\right]}_t}\\ \left(0.68s^{-1}\right)\times \left(5.0s\right)=\ln \frac{0.17M}{{\left[A\right]}_t}\\ 3.4=\ln \frac{0.17M}{{\left[A\right]}_t}\\ \frac{0.17M}{{\left[A\right]}_t}=e^{3.4}\\ {\left[A\right]}_t=\frac{0.17M}{e^{3.4}}\\ =5.6\times 10^{-3}M.\end{array}$

The concentration of azomethane remains is $5.6\times 10^{-3}M.$

Therefore the mass of azomethane remains can be calculated as follows,

    $\begin{array}{c}Molarity=\frac{\frac{mass}{malecularmass}}{volume}\\ \\ \text{5}{\text{.6×10}}^{\text{-3}}\text{M}=\frac{\frac{mass}{58.08g/mol}}{2L}\\ \\ mass=\left(\text{5}{\text{.6×10}}^{\text{-3}}\text{M}\right)\times \left(2L\right)\times \left(58.08g/mol\right)\\ \\ =0.65g.\end{array}$

The mass of azomethane remains after $5.0s$ is $0.65g.$

(c)

Interpretation Introduction

Interpretation:

The time taken for the azomethane to drop from $2gto0.24g$ has to be calculated.

Concept Introduction:

Refer to part (b).

Answer to Problem 100QRT

The time taken for the concentration to drop from $2gto0.24g$ is $3.09s.$

Explanation of Solution

Given,

  $\lambda =0.68s^{-1}$

    ${\left[A\right]}_o=\frac{\frac{2g}{58.08g/mol}}{2L}=0.017M$

    ${\left[A\right]}_t=\frac{\frac{0.24g}{58.08g/mol}}{2L}=2.066\times 10^{-3}M$

The time taken for the azomethane to drop from $2gto0.24g$ can be calculated as follows,

    $\begin{array}{c}t=\frac{1}{\lambda }\ln \frac{{\left[A\right]}_o}{{\left[A\right]}_t}\\ \\ =\frac{1}{0.68s^{-1}}\ln \left(\frac{\text{0}\text{.017}\text{M}}{\text{2}{\text{.066×10}}^{\text{-3}}\text{M}}\right)\\ \\ =3.09s.\end{array}$

The time taken for the concentration to drop from $2gto0.24g$ is $3.09s.$

(d)

Interpretation Introduction

Interpretation:

The mass of nitrogen found in the flask after $0.5s$ has to be calculated.

Concept Introduction:

Refer to part (b).

Answer to Problem 100QRT

The mass of nitrogen remaining in the flask after $0.5s$ is $0.3g{N}_2.$

Explanation of Solution

Mass of $C_2{H}_6{N}_2$ remained after $0.5s$ can be calculated using following equation.

  $t=\frac{1}{\lambda }\ln \frac{{\left[A\right]}_o}{{\left[A\right]}_t}$

Here mass is proportional to concentration.  So instead of concentration mass can be taken for calculation as follows,

    $\begin{array}{c}t=\frac{1}{\lambda }\ln \frac{{\left[A\right]}_o}{{\left[A\right]}_t}\\ \\ \ln {\left[A\right]}_t=-\lambda t+\ln {\left[A\right]}_o\\ \\ =\left(0.68s^{-1}\right)\left(0.5s\right)+\ln \left(2g\right)\\ \\ massof{\left[A\right]}_t=e^{-0.4}\\ \\ =1.4g.\end{array}$

$1.4g$ of $C_2{H}_6{N}_2$ is remaining in the system.

Amount of $C_2{H}_6{N}_2$ used can be calculated as follows,

  $\begin{array}{c}{C}_2{H}_6{N}_{2}used=initial{C}_2{H}_6{N}_2-final{C}_2{H}_6{N}_2\\ \\ =2g-1.4g\\ \\ =0.6g.\end{array}$

Hence mass of nitrogen remaining in the flask can be calculated as follows,

    $0.6g\times \frac{1mol{C}_2{H}_6{N}_2}{58.1g{C}_2{H}_6{N}_2}\times \frac{1mol{N}_2}{1mol{C}_2{H}_6{N}_2}\times \frac{28.02g{N}_2}{1mol{N}_2}=0.3g{N}_2.$

The mass of nitrogen remaining in the flask after $0.5s$ is $0.3g{N}_2.$