Chapter 6, Problem 6C.12E

(a)

Interpretation Introduction

Interpretation:

For the given first order reaction, the half-life period and the total pressure of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ at $\text{5}\text{.0}\text{s}$ have to be found.

Concept introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$.

Integrated rate law for first order reaction:

Consider A as substance, that gives the product based on the equation,

    $\text{aA}\to \text{products}$

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

    $\text{Rate=-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=k}\left[\text{A}\right]$

The integrated rate law equation can be given as,

    $\text{ln}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{o}}}\text{=-kt}$

The above expression is called integrated rate law for first order reaction.

Half-life period:

Half-life is defined as the time taken for amount of the sample to reduce to half of its starting value.  The symbol is ${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$.

$t_{\frac{1}{2}}=\frac{\ln \left(2\right)}{k}$

Half-life for a first order reaction is, $t_{\frac{1}{2}}=\frac{0.693}{k}$

Explanation of Solution

Given first-order decomposition reaction is,

$2N_2{O}_5(g)\underset{}{\to }4N{O}_2(g)+O_2(g)$

Half-life for a first order reaction is,

 $\begin{array}{l}{t}_{\frac{1}{2}}=\frac{\ln \left(2\right)}{k}\\ k=3.38\times {10}^{-5}{s}^{-1}at{25}^{o}C\end{array}$

Given value is substituted for above equation,

$\begin{array}{l}{t}_{\frac{1}{2}}=\frac{\ln \left(2\right)}{3.38\times {10}^{-5}{s}^{-1}}\\ t_{\frac{1}{2}}=\frac{0.6931}{3.38\times {10}^{-5}{s}^{-1}}=2.0502\times {10}^{-6}s\end{array}$

Therefore, the half-life of $N_2{O}_5$ is $2.0502\times {10}^{-6}s$

(b)

Interpretation Introduction

Interpretation:

For the given first order reaction, the total pressure of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ at $5.0$ min after the initiation of the reaction has to be found.

Concept introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\begin{array}{l}V\alpha \frac{nT}{P}\\ V=R\frac{nT}{P}\\ PV=nRT\end{array}$

Where,

 $\begin{array}{l}n=molesofgas\\ P=\text{Presure}\\ \text{T}\text{=Temperature}\\ \text{R=}\text{Gas}\text{constant}\end{array}$

Integrated rate law for first order reaction:

Consider A as substance, that gives the product based on the equation,

    $\text{aA}\to \text{products}$

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

    $\text{Rate=-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=k}\left[\text{A}\right]$

The integrated rate law equation can be given as,

    $\text{ln}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{o}}}\text{=-kt}$

The above expression is called integrated rate law for first order reaction.

Explanation of Solution

For a first order reaction $[A]={[A]}_0{e}^{-kt}$ (or) in the presence case if we assume the volume and temperature are held constant.

$P_{N_2{O}_5}=P_{N_2{O}_5}^{t=0}\times e^{-kt}$

The pressure is directly proportional to the number of moles $N_2{O}_5$

After 10 s, the pressure due to $N_2{O}_5$ will be reduced to  

$\begin{array}{l}{P}_{Initial}=78.4torr\\ PV=nRT\\ C=n/V=P/(RT)\\ C_{Initial}=P/(RT)\\ =78.4/62.4\times 298=4.216\times {10}^{-3}\end{array}$

Next we calculate the final pressure,

$\begin{array}{l}In(final)=In(C_{Initial})-k\times t\\ In(final)=In\left(4.216\times {10}^{-3}\right)-\left(3.38\times {10}^{-5}\right)\left(10\right)\\ In(final)=\left(-5.4688\right)-\left(3.38\times {10}^{-5}\right)\left(10\right)\\ C(final)=e^{-5.4691}\\ C(final)=4.215\times {10}^{-3}\end{array}$

Therefore, the final pressure is,

$\begin{array}{l}{P}_{Final}=C\times RT\\ P_{Final}=4.215\times {10}^{-3}\times 62.4\times 298\\ P_{Final}=78.378torr\end{array}$