Text book Example problems - Chapter 7 - Chemical Kinetics - 12th Chemistry Guide Samacheer Kalvi Solutions

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Text book Example problems
Question 1.
The oxidation of nitric oxide (NO)
$
2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{NO}_{2(\mathrm{~g})}
$
Series of experiments are conducted by keeping the concentration of one of the reactants constant and the changing the concentration of the others.

Find out the individual overall order of the reaction.
Solution:
Rate $=\mathrm{k}[\mathrm{NO}]^{\mathrm{m}}\left[\mathrm{O}_2\right]^{\mathrm{n}}$
For experiment 1, the rate law is
Rate $_2=\mathrm{k}[\mathrm{NO}]^{\mathrm{m}}\left[\mathrm{O}_2\right]^{\mathrm{n}}$
$19.26 \times 10^{-2}=\mathrm{k}[1.3]^{\mathrm{m}}[1.1]^{\mathrm{n}}$
Similarly for experiment 2
Rate $_2=\mathrm{k}[\mathrm{No}]^{\mathrm{m}}\left[\mathrm{O}_2\right]^{\mathrm{n}}$
$38.40 \times 10^{-2}=\mathrm{k}[1.3]^{\mathrm{m}}[2.2]^{\mathrm{n}}$
For experiment 3
Rate $_3=\mathrm{k}[\mathrm{NO}]^{\mathrm{m}}\left[\mathrm{O}_2\right]^{\mathrm{n}}$
$76.8 \times 10^{-2}=\mathrm{k}[2.6]^{\mathrm{m}}[1.1]^{\mathrm{n}}$
Dividing Eq (2) by Eq (1)
$\frac{38.40 \times 10^{-2}}{19.26 \times 10^{-2}}=\frac{k[1.3]^m[2.2]^n}{k[1.3]^m[1.1]^n}$
$2=\left(\frac{2.2}{1.1}\right)^n$
$2=2^{\mathrm{n}}$
i.e., $n=1$
Therefore the reaction is first order with respect to $\mathrm{O}_2$
Dividing Eq (3) byEq (1)
$\frac{76.8 \times 10^{-2}}{19.26 \times 10^{-2}}=\frac{k[2.6]^m[1.1]^n}{k[1.3]^m[1.1]^n}$
$4=\left(\frac{2.6}{1.3}\right)^m$
$4=2^{\mathrm{m}}$
i.e., $m=2$
Therefore the reaction is second order with respect to $\mathrm{NO}$
The rate law is Rate $_1=\mathrm{k}[\mathrm{NO}]^2\left[\mathrm{O}_2\right]^1$
The overall order of the reaction $=(2+1)=3$
Question 2.
Consider the oxidation of nitric oxide to form $\mathrm{NO}_2$
$2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}$

- Express the rate of the reaction in terms of changes in the concentration of $\mathrm{NO}, \mathrm{O}_2$ and $\mathrm{NO}_2$.
- At a particular instant, when $\left[\mathrm{O}_2\right]$ is decreasing at $0.2 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ at what rate is $\left[\mathrm{NO}_2\right]$ increasing at that instant?
Solution:
(a) Rate $=\frac{-1}{2} \frac{\mathrm{d}[\mathrm{NO}]}{\mathrm{dt}}=\frac{-\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}$
(b) $\frac{-\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}$
$
\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=2 \times\left(\frac{-\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}\right)=2 \times 0.2 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=0.4 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
$
Question 3.
What is the order with respect to each of the reactant and overall order of the following reactions?
1. $5 \mathrm{Br}^{-}{ }_{(\mathrm{aq})}+\mathrm{BrO}_3^{-}(\mathrm{aq})+6 \mathrm{H}_{(\mathrm{aq})}^{+} \rightarrow 3 \mathrm{Br}_{2(1)}+3 \mathrm{H}_2 \mathrm{O}_{(1)}$
The experimental rate law is
Rate $=\mathrm{k}\left[\mathrm{Br}^{-}\right]\left[\mathrm{BrO}_3{ }^{-}\right]\left[\mathrm{H}^{+}\right]^2$
2. $\mathrm{CH}_3 \mathrm{CHO}_{(\mathrm{g})} \triangle \mathrm{CH}_{4(\mathrm{~g})}+\mathrm{CO}_{(\mathrm{g})}$
the experimental rate law is
Rate $=\mathrm{k}\left[\mathrm{CH}_3 \mathrm{CHO}\right]^{3 / 2}$
Solution:
1. First order with respect to $\mathrm{Br}^{-}$, first order with respect to $\mathrm{BrO}_3{ }^{-}$and second order with respect to
$\mathrm{H}^{+}$. Hence the overall order of the reaction is equal to $1+1+2=4$
2. Order of the reaction with respect to acetaldehyde is $\frac{3}{2}$ and overall order is also $\frac{3}{2}$
Question 4.
The rate of the reaction $x+2 y \rightarrow$ product is $4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$, if $[x]=[y]=0.2 \mathrm{M}$ and rate constant at $400 \mathrm{~K}$ is $2 \times 10^{-2} \mathrm{~s}^{-1}$. what is the overall order of the reaction.
Solution:
$
\text { Rate }=\mathrm{k}[x]^{\mathrm{n}}[y]^{\mathrm{m}}
$

$
\begin{aligned}
& 4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=2 \times 10^{-2} \mathrm{~s}^{-1}\left(0.2 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{\mathrm{n}}\left(0.2 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{\mathrm{m}} \\
& \frac{4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}}{2 \times 10^{-2} \mathrm{~s}^{-1}}=(0.2)^{\mathrm{n}-\mathrm{m}}\left(\mathrm{mol} \mathrm{L}^{-1}\right)^{\mathrm{n}-\mathrm{m}} \\
& 2.0\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)=(0.2)^{\mathrm{n}+\mathrm{m}}\left(\mathrm{mol} \mathrm{L}^{-1}\right)^{\mathrm{n}-\mathrm{m}}
\end{aligned}
$
Comparing the powers on both sides, the overall order of the reaction $n+m=1$
Question 5.
A first order reaction takes 8 hours for $90 \%$ completion. Calculate the time required for $80 \%$ completion. $(\log 5=0.6989 ; \log 10=1)$
Solution:
For a first order reaction,
$
\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \left(\frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}\right)
$
Let $\left[\mathrm{A}_0\right]=100 \mathrm{M}$
When
$
\begin{aligned}
& \mathrm{t}=\mathrm{t}_{90 \%} ;[\mathrm{A}]=10 \mathrm{M} \text { (given that } \mathrm{t}_{90 \%}=8 \text { hours) } \\
& \mathrm{t}=\mathrm{t}_{80 \%} ;[\mathrm{A}]=20 \mathrm{M} \\
& \mathrm{k}=\frac{2.303}{\mathrm{t}_{80 \%}} \log \left(\frac{100}{20}\right) \\
& \mathrm{t}_{80 \%}=\frac{2.303}{\mathrm{k}} \log (5)
\end{aligned}
$
Find the value of $\mathrm{k}$ using the given data
$
\begin{aligned}
& \mathrm{k}=\frac{2.303}{t_{00 \%}} \log \left(\frac{100}{10}\right) \\
& \mathrm{k}=\frac{2.303}{8 \text { hours }} \log 10 \\
& \mathrm{k}=\frac{2.303}{8 \text { hours }}
\end{aligned}
$
Substitute the value of $\mathrm{k}$ in equation (2)
$
\begin{aligned}
& \mathrm{t}_{80 \%}=\frac{2.303}{2.303 / 8 \text { hours }} \log (5) \\
& \mathrm{t}_{80 \%}=8 \text { hours } \mathrm{x} 0.6989 \\
& \mathrm{t}_{80 \%}=5.59 \text { hours }
\end{aligned}
$

Question 6.
The half life of a first order reaction $\mathrm{x} \rightarrow$ products is $6.932 \times 10^4 \mathrm{~s}$ at $500 \mathrm{~K}$. What percentage of $\mathrm{x}$ would be decomposed on heating at $500 \mathrm{~K}$ for $100 \mathrm{~min} .\left(\mathrm{e}^{0.06}=1.06\right)$
Solution:
Given $t_{1 / 2}=0.6932 \times 10^4 \mathrm{~s}$
To solve: when $t=100 \mathrm{~min}$
$
\frac{\left[\mathrm{A}_0\right]-[\mathrm{A}]}{\left[\mathrm{A}_0\right]} \times 100=\text { ? }
$
We know that for a first order reaction, $\mathrm{t}_{1 / 2}=\frac{0.6932}{k}$
$
\begin{aligned}
& \mathrm{k}=\frac{0.6932}{6.932 \times 10^4} \Rightarrow \mathrm{k}=10^{-5} \mathrm{~s}^{-1} \\
& \mathrm{k}=\left(\frac{1}{\mathrm{t}}\right) \ln \left(\frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}\right) \\
& 10^{-5} \mathrm{~s}^{-1} \times 100 \times 60 \mathrm{~s}=\ln \left(\frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}\right) \\
& 0.06=\ln \left(\frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}\right) \\
& \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}=e^{0.06}=1.06 \\
& \therefore \frac{\left[\mathrm{A}_0\right]-[\mathrm{A}]}{\left[\mathrm{A}_0\right]} \times 100 \%=\left(1-\frac{[\mathrm{A}]}{\left[\mathrm{A}_0\right]}\right) \times 100 \%=\left(1-\frac{1}{1.06}\right) \times 100 \%=5.6 \%
\end{aligned}
$
Question 7.
Show that in case of first order reaction, the time required for $99.9 \%$ completion is nearly ten times the time required for half completion of the reaction.
Solution:
Let $\left[\mathrm{A}_0\right]=100$

When $t=\mathrm{t}_{99.9 \%} ;[\mathrm{A}]=(100-99.9)=0.1$
$
\begin{aligned}
& \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \left(\frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}\right) \\
& \mathrm{t}_{99.9 \%}=\frac{2.303}{\mathrm{k}} \log \left(\frac{100}{0.1}\right) \Rightarrow \mathrm{t}_{99.9 \%}=\frac{2.303}{\mathrm{k}} \log 1000 \\
& \mathrm{t}_{99.9 \%}=\frac{2.303}{\mathrm{k}}(3) \Rightarrow \mathrm{t}_{99.9 \%}=\frac{6.906}{\mathrm{k}} \\
& \mathrm{t}_{99.9 \%} \simeq 10 \times \frac{0.69}{\mathrm{k}} \\
& \mathrm{t}_{99.9 \%} \simeq 10 \mathrm{t}_{1 / 2}
\end{aligned}
$
Question 8.
The rate constant of a reaction at 400 and $200 \mathrm{~K}$ are 0.04 and $0.02 \mathrm{~s}^{-1}$ respectively. Calculate the value of activation energy.
Answer:
According to Arrhenius equation.
$
\begin{aligned}
& \log \left(\frac{\mathrm{k}_2}{\mathrm{k}_1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_{\mathrm{i}} \mathrm{T}_2}\right) \\
& \mathrm{T}_2=400 \mathrm{~K} ; \mathrm{k}_2=0.04 \mathrm{~s}^{-1} \\
& \mathrm{~T}_1=200 \mathrm{~K} ; \mathrm{k}_1=0.02 \mathrm{~s}^{-1} \\
& \log \left(\frac{0.04 \mathrm{~s}^{-1}}{0.02 \mathrm{~s}^{-1}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}\left(\frac{400 \mathrm{~K}-200 \mathrm{~K}}{200 \mathrm{~K} \times 400 \mathrm{~K}}\right) \\
& \log (2)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}\left(\frac{1}{400 \mathrm{~K}}\right) \\
& \mathrm{E}_{\mathrm{a}}=\log (2) \times 2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 400 \mathrm{~K} \\
& \mathrm{E}_{\mathrm{a}}^{-}=2305 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 9.
Rate constant $\mathrm{k}$ of a reaction varies with temperature $\mathrm{T}$ according to the following Arrhenius equation. $\log \mathrm{k}=\log \mathrm{A} \frac{E_a}{2.303 R} \frac{1}{T}$.

Where $\mathrm{E}$ is the activation energy. When a graph is plotted for $\log \mathrm{k} V \mathrm{~s} \frac{1}{T}$ a straight line with a slope of $4000 \mathrm{~K}$ is obtained. Calculate the activation energy.
Solution:
$
\begin{aligned}
& \log \mathrm{k}=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{\mathrm{~T}}\right) \\
& \mathrm{y}=\mathrm{c}+\mathrm{mx} \\
& m=-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}} \\
& \mathrm{E}_{\mathrm{a}}=-2.303 \mathrm{Rm} \\
& \mathrm{E}_{\mathrm{a}}=-2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times(-4000 \mathrm{k}) \\
& \mathrm{E}_{\mathrm{a}}=76,589 \mathrm{~J} \mathrm{~mol}^{-1} \\
& \mathrm{E}_{\mathrm{a}}=76,589 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$