Chapter 6, Problem 6C.5E

(a)

Interpretation Introduction

Interpretation:

The rate constant of the zeroth-order reaction has to be calculated.

Concept introduction:

Integrated law for zero-order reactions:

Consider a reaction,

    $\text{aA}\to \text{products}$

Then, the rate of the reaction is

    $\text{Rate=k}{\left[\text{A}\right]}^{\text{0}}$

The expression for this reaction can be written as,

    $\text{Rate=k}$

According to this law, the rate would not change with change in concentration.

The relationship between concentration and time for a reaction that is zero-order is given as,

    ${\left[\text{A}\right]}_{\text{t}}\text{=-kt+}{\left[\text{A}\right]}_{\text{0}}$

Explanation of Solution

Rate constant for given zero order reaction,

$\begin{array}{l}A\stackrel{}{\to }B\\ -rA=-d[A]dt=k[A]\\ \Rightarrow -d[A]dt=k\\ \Rightarrow d[A]=-kdt\end{array}$

Integration the above equation,

$\begin{array}{l}\Rightarrow \int d[A]=-k\int dt\\ [A]=-kt+c-----[1]\\ \text{'}c\text{'}is\text{integration}\text{constant}\end{array}$

At time $t=0$

$\begin{array}{l}[A]={\left[A\right]}^0\\ Fromequation(1)\\ {\left[A\right]}^0=C\end{array}$

From equation (1)

$[A]=-kt+c$

This equation is zero order reaction, and given values is substituted for this equation and calculates ‘k’

 $\begin{array}{l}10=-k\times 770+21\\ -k\times 770+21=10\\ subtract21frombothsides\\ -k\times 770+21-21=10-21\\ simplify\\ k\times 770=-11\\ Dividebothsidesby-770\\ \frac{-k\times 770}{-770}=\frac{-11}{-770}\\ k=0.01429\end{array}$

 Therefore, the rate constant value is $k=0.01429\text{M/S}$

(b)

Interpretation Introduction

Interpretation:

The time taken for the disappearance of all ammonia has to be found.

Concept introduction:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how much concentration of targets (products) formed in a unit of time is said to be rate of reaction.

Rate of reaction depends on time, temperature, pressure, concentration, and $\text{pH}$ of the reaction.

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area…

General rate reaction is,

$aA+bB\to cC+dD$

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

The negative sign indicates the reduction of concentration of reactant.

$\Delta \left[A\right],\Delta \left[B\right],\Delta \left[C\right]and\Delta \left[D\right]$ are the change in concentration of A,B, C and D over time period $\Delta t$

Explanation of Solution

Integrated rate equation for zero order reaction,

$\begin{array}{l}[A]={\left[A\right]}^0-kt\\ -kt={\left[A\right]}^0-[A]\\ k=\frac{{[A]}^0-[A]}{t}\end{array}$

${[A]}^0=$ Initial pressure $21kPa$

$[A]=$ Final partial pressure

Hence,

$\begin{array}{l}k=\frac{21-10}{770}\\ k=\frac{11}{770}\\ k=0.0142s^{-1}\end{array}$

When all ammonia is consumed (or) disappears is $[A]=0$

$\begin{array}{l}t=\frac{[A_0]-[A]}{k}\\ t=\frac{21}{0.0142}\\ t=1500s\end{array}$