Answer

(i) Given rate $=k[\mathrm{NO}]^{2}$

Therefore, order of the reaction $=2$

Dimension of $k=\frac{\text { Rate }}{[\mathrm{NO}]^{2}}$

$=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}}$

$=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol}^{2} \mathrm{~L}^{-2}}$

$=\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$

(ii) Given rate $=k\left[\mathrm{H_2} \mathrm{O_2}\right]\left[\mathrm{l}^{-}\right]$

Therefore, order of the reaction $=2$

Dimension of $ k=\frac{\text { Rate }}{\left[\mathrm{H_2} \mathrm{O_2}\right]\left[\mathrm{I}^{-}\right]} $

$ \begin{aligned} & =\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)\left(\mathrm{mol} \mathrm{L}^{-1}\right)} \\ & =\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1} \end{aligned} $

(iii) Given rate $=k\left[\mathrm{CH_3} \mathrm{CHO}\right]^{3 / 2}$

Therefore, order of reaction $=\frac{3}{2}$

Dimension of $ k=\frac{\text { Rate }}{\left[\mathrm{CH_3} \mathrm{CHO}\right]^{\frac{3}{2}}} $

$ \begin{aligned} & =\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^{\frac{3}{2}}} \\ & =\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol}^{\frac{3}{2}} \mathrm{~L}^{-\frac{3}{2}}} \\ & =\mathrm{L}^{\frac{1}{2}} \mathrm{~mol}^{-\frac{1}{2}} \mathrm{~s}^{-1} \end{aligned} $

(iv) Given rate $=k\left[\mathrm{C_2} \mathrm{H_5} \mathrm{Cl}\right]$

Therefore,order of the reaction $=1$

Dimension of $k=\frac{\text { Rate }}{\left[\mathrm{C_2} \mathrm{H_5} \mathrm{Cl}\right]}$

$=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol} \mathrm{~L}^{-1}}$

$=\mathrm{s}^{-1}$

Answer

The initial rate of the reactionis

Rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$

$=\left(2.0 \times 10^{-6} \mathrm{~mol}^{-2} \mathrm{~L}^{2} \mathrm{~S}^{-1}\right)\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.2 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}$

$=8.0 \times 10^{-9} \mathrm{~mol}^{-2} \mathrm{~L}^{2} \mathrm{~S}^{-1}$

When $[A]$ is reduced from $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$ to $0.06 \mathrm{~mol}^{-1}$, the concentration of A reacted $=(0.1-0.06) \mathrm{mol} \mathrm{L}^{-1}=0.04 \mathrm{~mol} \mathrm{~L}^{-1}$

Therefore, concentration of B reacted $=\frac{1}{2} \times 0.04 \mathrm{~mol} \mathrm{~L}^{-1}=0.02 \mathrm{~mol} \mathrm{~L}^{-1}$

Then, concentration of $B$ available, $[B]=(0.2-0.02) \mathrm{mol} \mathrm{L}^{-1}$

$=0.18 \mathrm{~mol} \mathrm{~L}^{-1}$

After $[A]$ is reduced to $0.06 \mathrm{~mol} \mathrm{~L}^{-1}$, the rate of the reaction is given by,

Rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$

$=\left(2.0 \times 10^{-6} \mathrm{~mol}^{-2} \mathrm{~L}^{2} \mathrm{~S}^{-1}\right)\left(0.06 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.18 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}$ $=3.89 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~S}^{-1}$

Answer

The decomposition of $\mathrm{NH_3}$ on platinum surface is represented by the following equation.

$2 \mathrm{NH_3(g)} \xrightarrow{\mathrm{Pt}} \mathrm{N_2(g)}+3 \mathrm{H_2(g)}$

Therefore,

Rate $=-\frac{1}{2} \frac{d\left[\mathrm{NH_3}\right]}{d t}=\frac{d\left[\mathrm{~N_2}\right]}{d t}=\frac{1}{3} \frac{d\left[\mathrm{H_2}\right]}{d t}$

However, it is given that the reaction is of zero order.

Therefore,

$$ \begin{aligned} -\frac{1}{2} \frac{d\left[\mathrm{NH_3}\right]}{d t}=\frac{d\left[\mathrm{~N_2}\right]}{d t}=\frac{1}{3} \frac{d\left[\mathrm{H_2}\right]}{d t} & =k \\ & =2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \end{aligned} $$

Therefore, the rate of production of $\mathrm{N_2}$ is

$$ \frac{d\left[\mathrm{~N_2}\right]}{d t}=2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} $$

And, the rate of production of $\mathrm{H_2}$ is

$$ \frac{d\left[\mathrm{H_2}\right]}{d t}=3 \times 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} $$

$=7.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~S}^{-1}$

Answer

If pressure is measured in bar and time in minutes, then

Unit of rate $=$ bar $\min ^{-1}$

Rate $=k\left(p_{\mathrm{CH_3} \mathrm{OCH_3}}\right)^{3 / 2}$

$\Rightarrow k=\frac{\text { Rate }}{\left(p_{\mathrm{CH_3} \mathrm{OCH_3}}\right)^{3 / 2}}$

Therefore, unit of rate constants $(k) = \frac{\text{bar min}^{-1}}{\text{bar}^{3/2}} $

$=\operatorname{bar}^{-1 / 2} \min ^{-1}$

Answer

The factors that affect the rate of a reaction areas follows.

(i) Concentration of reactants (pressure in case of gases)

(ii) Temperature

(iii) Presence of a catalyst

Answer

Letthe concentration of the reactant be $[\mathrm{A}]=\mathrm{a}$

Rate of reaction, $\mathrm{R}=k[\mathrm{~A}]^{2}$

$=k a^{2}$

(i)If the concentration of the reactant is doubled, i.e. $[A]=2 a$, then the rate of the reaction would be

$ \mathrm{R}^{\prime}=k(2 a)^{2} $

$=4 \mathrm{ka}^{2}$

$=4 \mathrm{R}$

Therefore, the rate of the reaction would increase by 4 times.

(ii) If the concentration of the reactant is reduced to half, i.e.$[\mathrm{A}]=\frac{1}{2} a$, then the reate of the reaction would be

$ \begin{aligned} & R^{*}=k\left(\frac{1}{2} a\right)^{2} \\ & =\frac{1}{4} k a^{2} \\ & =\frac{1}{4} R \end{aligned} $

Therefore, the rate of the reaction would be reduced to $\frac{1}{4}^{\text {th }}$.

Answer

The rate constant is nearly doubled with a rise in temperature by $10^{\circ}$ for a chemical reaction.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,

$k=\mathrm{A} e^{-E_{\mathrm{s}} / \mathrm{R} T}$

where, kis the rate constant,

$A$ is the Arrhenius factor or the frequency factor,

$\mathrm{R}$ is the gas constant,

Tis the temperature, and

$E_{a}$ is the energy of activation for the reaction

Answer

(i) Average rate of reaction between the time interval, 30 to 60 seconds, $ =\frac{d[\text { Ester }]}{d t} $

$ \begin{aligned} & =\frac{0.31-0.17}{60-30} \\ & =\frac{0.14}{30} \\ & =4.67 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \end{aligned} $

(ii) For a pseudo first order reaction,

$k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$

For $t=30 \mathrm{~s}, \quad k_{1}=\frac{2.303}{30} \log \frac{0.55}{0.31}$

$=1.911 \times 10^{-2} \mathrm{~s}^{-1}$

For $t=60 \mathrm{~s}, \quad k_{2}=\frac{2.303}{60} \log \frac{0.55}{0.17}$

$=1.957 \times 10^{-2} \mathrm{~S}^{-1}$

For $t=90 \mathrm{~s}, \quad k_{3}=\frac{2.303}{90} \log \frac{0.55}{0.085}$

$=2.075 \times 10^{-2} \mathrm{~S}^{-1}$

Then, average rate constant,
$ k=\frac{k_{1}+k_{2}+k_{3}}{3} $

$ \begin{aligned} & =\frac{\left(1.911 \times 10^{-2}\right)+\left(1.957 \times 10^{-2}\right)+\left(2.075 \times 10^{-2}\right)}{3} \\ & =1.98 \times 10^{-2} \mathrm{~s}^{-1} \end{aligned} $

Answer

(i) The differential rate equation will be

$ -\frac{d[\mathrm{R}]}{d t}=k[\mathrm{~A}][\mathrm{B}]^{2} $

(ii) If the concentration of $\mathrm{B}$ is increased three times, then

$ \begin{aligned} -\frac{d[\mathrm{R}]}{d t} & =k[\mathrm{~A}][3 \mathrm{~B}]^{2} \\ & =9 \cdot k[\mathrm{~A}][\mathrm{B}]^{2} \end{aligned} $

Therefore, the rate of reaction will increase 9 times.

(iii) When the concentrations of both $A$ and $B$ are doubled,

$ \begin{aligned} -\frac{d[\mathrm{R}]}{d t} & =k[\mathrm{~A}][\mathrm{B}]^{2} \\ & =k[2 \mathrm{~A}][2 \mathrm{~B}]^{2} \\ & =8 \cdot k[\mathrm{~A}][\mathrm{B}]^{2} \end{aligned} $

Therefore, the rate of reaction will increase 8 times.

Answer

Let the order of the reaction with respect to $\mathrm{A}$ be xand with respect to $\mathrm{B}$ be $y$.

Therefore, $\mathrm{r_0}=k[\mathrm{~A}]^{x}[\mathrm{~B}]^{y}$

$5.07 \times 10^{-5}=k[0.20]^{x}[0.30]^{v}\quad \quad \quad \text{(i)}$

$5.07 \times 10^{-5}=k[0.20]^{x}[0.10]^{y}\quad \quad \quad \text{(ii)}$

$1.43 \times 10^{-4}=k[0.40]^{x}[0.05]^{y}\quad \quad \quad \text{(iii)}$

Dividing equation (i) by (ii), we obtain

$\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}}=\frac{k[0.20]^{x}[0.30]^{y}}{k[0.20]^{x}[0.10]^{y}}$

$\Rightarrow 1=\frac{[0.30]^{y}}{[0.10]^{y}}$

$\Rightarrow\left(\frac{0.30}{0.10}\right)^{0}=\left(\frac{0.30}{0.10}\right)^{y}$

$\Rightarrow y=0$

Dividing equation (iii) by (ii), we obtain

$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{k[0.40]^{x}[0.05]^{y}}{k[0.20]^{x}[0.30]^{y}}$

$\Rightarrow \frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{[0.40]^{x}}{[0.20]^{x}} \quad\left[\begin{array}{l}\text { Since } y=0, \\ {[0.05]^{y}=[0.30]^{y}=1}\end{array}\right]$

$\Rightarrow 2.821=2^{x}$

$\Rightarrow \log 2.821=x \log 2 \quad$ (Taking log on both sides)

$\Rightarrow x=\frac{\log 2.821}{\log 2}$

$=1.496$

$=1.5$ (approximately)

Hence, the order of the reaction with respect to $A$ is 1.5 and with respect to $B$ is zero.

Answer

Let the order of the reaction with respect to $\mathrm{A}$ be xand with respect to $\mathrm{B}$ be $y$.

Therefore, rate of the reaction is given by,

Rate $=k[\mathrm{~A}]^{x}[\mathrm{~B}]^{y}$

According to the question,

$6.0 \times 10^{-3}=k[0.1]^{x}[0.1]^{y}\quad \quad \quad \text{(i)}$

$7.2 \times 10^{-2}=k[0.3]^{x}[0.2]^{y}\quad \quad \quad \text{(ii)}$

$2.88 \times 10^{-1}=k[0.3]^{x}[0.4]^{y}\quad \quad \quad \text{(iii)}$

$2.40 \times 10^{-2}=k[0.4]^{x}[0.1]^{y}\quad \quad \quad \text{(iv)}$

Dividing equation (iv) by (i), we obtain

$ \begin{aligned} & \frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}}=\frac{k[0.4]^{x}[0.1]^{y}}{k[0.1]^{x}[0.1]^{y}} \\ & \Rightarrow 4=\frac{[0.4]^{x}}{[0.1]^{x}} \\ & \Rightarrow 4=\left(\frac{0.4}{0.1}\right)^{x} \\ & \Rightarrow(4)^{1}=4^{x} \\ & \Rightarrow x=1 \end{aligned} $

Dividing equation (iii) by (ii), we obtain

Therefore, the rate law is

Rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$

$\Rightarrow k=\frac{\text { Rate }}{[\mathrm{A}][\mathrm{B}]^{2}}$

From experiment I, we obtain

$k=\frac{6.0 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}}$

$=6.0 \mathrm{~L}^{2} \mathrm{~mol}^{-2} \mathrm{~min}^{-1}$

From experiment II, weobtain

$k=\frac{7.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.3 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.2 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}}$

$=6.0 \mathrm{~L}^{2} \mathrm{~mol}^{-2} \mathrm{~min}^{-1}$

From experiment III, we obtain

$k=\frac{2.88 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.3 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}}$

$=6.0 \mathrm{~L}^{2} \mathrm{~mol}^{-2} \mathrm{~min}^{-1}$

From experiment IV, we obtain

$k=\frac{2.40 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}}$

$=6.0 \mathrm{~L}^{2} \mathrm{~mol}^{-2} \mathrm{~min}^{-1}$

Therefore, rate constant, $k=6.0 \mathrm{~L}^{2} \mathrm{~mol}^{-2} \mathrm{~min}^{-1}$

Answer

The given reaction is of the first order with respect to $A$ and of zero order with respect to $B$.

Therefore, the rate of the reaction is given by,

Rate $=k[\mathrm{~A}]^{1}[\mathrm{~B}]^{0}$

$\Rightarrow$ Rate $=k[A]$

From experiment I, we obtain

$2.0 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}=\mathrm{k}\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)$

$\Rightarrow k=0.2 \mathrm{~min}^{-1}$

From experiment II, we obtain

$4.0 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}=0.2 \mathrm{~min}^{-1}[\mathrm{~A}]$

$\Rightarrow[\mathrm{A}]=0.2 \mathrm{~mol} \mathrm{~L}^{-1}$

From experiment III, we obtain

Rate $=0.2 \mathrm{~min}^{-1} \times 0.4 \mathrm{~mol} \mathrm{~L}^{-1}$

$=0.08 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$

From experiment IV, we obtain

$2.0 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}=0.2 \mathrm{~min}^{-1}[\mathrm{~A}]$

$\Rightarrow[A]=0.1 \mathrm{~mol} \mathrm{~L}^{-1}$

Answer

(i) Half life, $t_{1 / 2}=\frac{0.693}{k}$

$=\frac{0.693}{200 \mathrm{~s}^{-1}}$

$=3.47 \times 10^{-3} \mathrm{~s}$ (approximately)

(ii) Half life, $t_{1 / 2}=\frac{0.693}{k}$

$=\frac{0.693}{2 \min ^{-1}}$

$=0.35 \mathrm{~min}$ (approximately) (iii) Half life, $t_{1 / 2}=\frac{0.693}{k}$

$=\frac{0.693}{4 \text { years }^{-1}}$

$=0.173$ years (approximately)

Answer

Here, $k=\frac{0.693}{t_{1 / 2}}$

$=\frac{0.693}{5730}$ years $^{-1}$

It is known that,

$ \begin{aligned} t & =\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \\ & =\frac{2.303}{\frac{0.693}{5730}} \log \frac{100}{80} \end{aligned} $

$=1845$ years (approximately)

Hence, the age of the sample is 1845 years.

Answer

$\mathrm{t}(\mathrm{s}) \longrightarrow$

(ii) Time corresponding to the concentration, $\frac{1.630 \times 10^{2}}{2} \mathrm{~mol} \mathrm{~L}^{-1}=81.5 \mathrm{~mol} \mathrm{~L}^{-1}$, is the half life. From the graph, the half life is obtained as $1450 \mathrm{~s}$.

(iii)

(iv) The given reaction is of the first order as the plot, $\log \left[\mathrm{N_2} \mathrm{O_5}\right] \mathrm{v} / \mathrm{s} t$, is a straight line. Therefore, the rate law of the reaction is

Rate $=k\left[\mathrm{~N_2} \mathrm{O_5}\right]$

(v) From the plot, $\log \left[\mathrm{N_2} \mathrm{O_5}\right]$ v/s $t$, we obtain

$ \begin{aligned} \text { Slope } & =\frac{-2.46-(-1.79)}{3200-0} \\ & =\frac{-0.67}{3200} \end{aligned} $

Again, slope of the line of the plot $\log \left[\mathrm{N_2} \mathrm{O_5}\right]$ v/s $t$ is given by

$ -\frac{k}{2.303} $

Therefore, we obtain,

$ -\frac{k}{2.303}=-\frac{0.67}{3200} $

Answer

It is known that,

$ \begin{aligned} t & =\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \\ & =\frac{2.303}{60 \mathrm{~s}^{-1}} \log \frac{1}{1 / 16} \\ & =\frac{2.303}{60 \mathrm{~s}^{-1}} \log 16 \\ & =4.6 \times 10^{-2} \mathrm{~s} \text { (approximately) } \end{aligned} $

Hence, the required time is $4.6 \times 10^{-2} \mathrm{~s}$.

Answer

Here,

$ k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{28.1} \mathrm{y}^{-1} $

It is known that,

$ \begin{aligned} & t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \\ & \Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[\mathrm{R}]} \\ & \Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}}(-\log [\mathrm{R}]) \\ & \Rightarrow \log [\mathrm{R}]=-\frac{10 \times 0.693}{2.303 \times 28.1} \\ & \Rightarrow[\mathrm{R}]=\operatorname{antilog}(-0.1071) \\ & \quad=\operatorname{antilog}(\overline{1} .8929) \\ & \quad=0.7814 \mu \mathrm{g} \end{aligned} $

Therefore, $0.7814 \mu \mathrm{~g}$ of ${ }^{90} \mathrm{Sr}$ will remain after 10 years.

Again,

$ \begin{aligned} & t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \\ & \Rightarrow 60=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[\mathrm{R}]} \\ & \Rightarrow \log [\mathrm{R}]=-\frac{60 \times 0.693}{2.303 \times 28.1} \\ & \Rightarrow[\mathrm{R}]=\operatorname{antilog}(-0.6425) \\ & \quad=\operatorname{antilog}(\overline{1} .3575) \\ & \quad=0.2278 \mu \mathrm{g} \end{aligned} $

Therefore, $0.2278 \mu \mathrm{~g} $ of $ { }^{90} \mathrm{Sr}$ will remain after 60 years.

Answer

For a first order reaction, the time required for $99 \%$ completionis

$ \begin{aligned} t_{1} & =\frac{2.303}{k} \log \frac{100}{100-99} \\ & =\frac{2.303}{k} \log 100 \\ & =2 \times \frac{2.303}{k} \end{aligned} $

For a first order reaction, the time required for $90 \%$ completion is

$ \begin{aligned} t_{2} & =\frac{2.303}{k} \log \frac{100}{100-90} \\ & =\frac{2.303}{k} \log 10 \\ & =\frac{2.303}{k} \end{aligned} $

Therefore, $t_{1}=2 t_{2}$

Hence, the time required for $99 \%$ completion of a first order reaction is twice the time required for the completion of $90 \%$ of the reaction.

Answer

For a first order reaction,

$ \begin{aligned} t & =\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \\ k & =\frac{2.303}{40 \mathrm{~min}} \log \frac{100}{100-30} \\ & =\frac{2.303}{40 \mathrm{~min}} \log \frac{10}{7} \\ & =8.918 \times 10^{-3} \mathrm{~min}^{-1} \end{aligned} $

Therefore, $t_{1 / 2}$ of the decomposition reaction is

$ \begin{aligned} t_{1 / 2} & =\frac{0.693}{k} \\ & =\frac{0.693}{8.918 \times 10^{-3}} \mathrm{~min} \end{aligned} $

$=77.7 \mathrm{~min}$ (approximately)

Answer

The decomposition of azoisopropane to hexane and nitrogen at $543 \mathrm{~K}$ is represented by the following equation.

After time, $t$, total pressure, $\mathrm{P_t}=\left(\mathrm{P_0}-p\right)+p+p$

$\Rightarrow \mathrm{P_t}=\mathrm{P_0}+p$

$\Rightarrow p=\mathrm{P_\mathrm{t}}-\mathrm{P_0}$

Therefore, $\mathrm{P_\mathrm{o}}-p=\mathrm{P_\mathrm{o}}-\left(\mathrm{P_\mathrm{t}}-\mathrm{P_\mathrm{o}}\right)$

$=2 \mathrm{P_0}-\mathrm{P_t}$

For a first order reaction,

$k=\frac{2.303}{t} \log \frac{\mathrm{P_0}}{\mathrm{P_0}-p}$

$=\frac{2.303}{t} \log \frac{\mathrm{P_0}}{2 \mathrm{P_0}-\mathrm{P_t}}$

When $t=360 \mathrm{~s}, \quad k=\frac{2.303}{360 \mathrm{~s}} \log \frac{35.0}{2 \times 35.0-54.0}$

$=2.175 \times 10^{-3} \mathrm{~S}^{-1}$

When $t=720 \mathrm{~s}, \quad k=\frac{2.303}{720 \mathrm{~s}} \log \frac{35.0}{2 \times 35.0-63.0}$

$=2.235 \times 10^{-3} \mathrm{~S}^{-1}$

Hence, the average value of rate constant is

$k=\frac{\left(2.175 \times 10^{-3}\right)+\left(2.235 \times 10^{-3}\right)}{2} \mathrm{~s}^{-1}$

$=2.21 \times 10^{-3} \mathrm{~S}^{-1}$

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

Answer

The thermal decomposition of $\mathrm{SO_2} \mathrm{Cl_2}$ at a constant volume is represented by the following equation.

$\begin{array}{lllcc} & \mathrm{SO_2} \mathrm{Cl_{2(g)}} & \longrightarrow & \mathrm{SO_{2(g)}} &+\mathrm{Cl_{2(g)}} \\ \text { At } t=0 & \mathrm{P_0} & & 0 & 0 \\ \text { At } t=t & \mathrm{P_0}-\mathrm{p} & & \mathrm{p} & \mathrm{p} \end{array}$

After time, $t$, total pressure, $\mathrm{P_t}=\left(\mathrm{P_0}-p\right)+p+p$

$\Rightarrow \mathrm{P_t}=\mathrm{P_0}+p$

$\Rightarrow p=\mathrm{P_\mathrm{t}}-\mathrm{P_0}$

Therefore, $\mathrm{P_\mathrm{o}}-p=\mathrm{P_\mathrm{o}}-\left(\mathrm{P_\mathrm{t}}-\mathrm{P_\mathrm{o}}\right)$

$=2 P_{0}-P_{t}$

For a first order reaction,

$\begin{aligned} k & =\frac{2.303}{t} \log \frac{\mathrm{P_0}}{\mathrm{P_0}-p} \\ & =\frac{2.303}{t} \log \frac{\mathrm{P_0}}{2 \mathrm{P_0}-\mathrm{P_t}}\end{aligned}$

When $t=100 \mathrm{~s}, \quad k=\frac{2.303}{100 \mathrm{~s}} \log \frac{0.5}{2 \times 0.5-0.6}$

$=2.231 \times 10^{-3} \mathrm{~S}^{-1}$

When $\mathrm{P_t}=0.65 \mathrm{~atm}$,

$\mathrm{P_0}+p=0.65$

$\Rightarrow p=0.65-P_{0}$

$=0.65-0.5$

$=0.15 \mathrm{~atm}$

Therefore, when the total pressure is $0.65 \mathrm{~atm}$, pressure of $\mathrm{SOCl_2}$ is

$p_{\mathrm{SOCl_2}}=\mathrm{P_0}-\mathrm{p}$

$=0.5-0.15$

$=0.35 \mathrm{~atm}$

Therefore, the rate of equation, when total pressure is $0.65 \mathrm{~atm}$, is given by,

Rate $=k\left({ }^{p_{\mathrm{SOCl_2}}}\right)$

$=\left(2.23 \times 10^{-3} \mathrm{~S}^{-1}\right)(0.35 \mathrm{~atm})$

$=7.8 \times 10^{-4} \mathrm{~atm} \mathrm{~s} \mathrm{~s}^{-1}$

Answer

From the given data, we obtain

Slope of the line,

$$ \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=-12.301 \mathrm{~K} $$

According to Arrhenius equation,

Slope $=-\frac{E_{a}}{\mathrm{R}}$

$\Rightarrow E_{a}=-$ Slope $\times \mathrm{R}$

$=-(-12.301 \mathrm{~K}) \times\left(8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$

$=102.27 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Again,

$\ln k=\ln A-\frac{E_{a}}{\mathrm{R} T}$

$\ln A=\ln k+\frac{E_{a}}{\mathrm{R} T}$

When $T=273 \mathrm{~K}$,

$\ln k=-7.147$

Then, $\ln A=-7.147+\frac{102.27 \times 10^{3}}{8.314 \times 273}$

$$ =37.911 $$

Therefore, $A=2.91 \times 10^{6}$

When $T=30+273 \mathrm{~K}=303 \mathrm{~K}$,

$\frac{1}{T}=0.0033 \mathrm{~K}=3.3 \times 10^{-3} \mathrm{~K}$

Then,

$$ \text { at } \frac{1}{T}=3.3 \times 10^{-3} \mathrm{~K} \text {, } $$

$\ln k=-2.8$

Therefore, $k=6.08 \times 10^{-2} \mathrm{~s}^{-1}$

Again, when $T=50+273 \mathrm{~K}=323 \mathrm{~K}$,

Answer

$k=2.418 \times 10^{-5} \mathrm{~S}^{-1}$

$T=546 \mathrm{~K}$ $E_{\mathrm{a}}=179.9 \mathrm{~kJ} \mathrm{~mol}^{-1}=179.9 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}$

According to the Arrhenius equation,

$k=\mathrm{Ae}^{-E_{\mathrm{a}} / \mathrm{R} T}$

$\Rightarrow \ln k=\ln \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$

$\Rightarrow \log k=\log \mathrm{A}-\frac{E_{a}}{2.303 \mathrm{R} T}$

$\Rightarrow \log \mathrm{A}=\log k+\frac{E_{a}}{2.303 \mathrm{R} T}$

$=\log \left(2.418 \times 10^{-5} \mathrm{~s}^{-1}\right)+\frac{179.9 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1} \times 546 \mathrm{~K}}$

$=(0.3835-5)+17.2082$

$=12.5917$

Therefore, $\mathrm{A}=\operatorname{antilog}$ (12.5917)

$=3.9 \times 10^{12} \mathrm{~S}^{-1}$ (approximately)

Answer

$k=2.0 \times 10^{-2} \mathrm{~s}^{-1}$

$T=100 \mathrm{~s}$

$[A]_{0}=1.0 \mathrm{moL}^{-1}$

Sincethe unit of $\mathrm{kis} \mathrm{s}^{-1}$, the given reaction is a first order reaction.

Therefore, $ k=\frac{2.303}{t} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]} $

$\Rightarrow 2.0 \times 10^{-2} \mathrm{~s}^{-1}=\frac{2.303}{100 \mathrm{~s}} \log \frac{1.0}{[\mathrm{~A}]}$

$\Rightarrow 2.0 \times 10^{-2} \mathrm{~s}^{-1}=\frac{2.303}{100 \mathrm{~s}}(-\log [\mathrm{A}])$

$\Rightarrow-\log [\mathrm{A}]=\frac{2.0 \times 10^{-2} \times 100}{2.303}$

$\Rightarrow[\mathrm{A}]=\operatorname{anti} \log \left(-\frac{2.0 \times 10^{-2} \times 100}{2.303}\right)$

$=0.135 \mathrm{~mol} \mathrm{~L}^{-1}$ (approximately)

Hence, the remaining concentration of $A$ is $0.135 \mathrm{~mol} \mathrm{~L}^{-1}$.

Answer

For a first order reaction,

$k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$

It is given that, $t_{1 / 2}=3.00$ hours

$ k=\frac{0.693}{t_{1 / 2}} $

Therefore,

$=\frac{0.693}{3} \mathrm{~h}^{-1}$

$=0.231 \mathrm{~h}^{-1}$

Then, $0.231 \mathrm{~h}^{-1}$

$ =\frac{2.303}{8 \mathrm{~h}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} $

$\Rightarrow \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}=\frac{0.231 \mathrm{~h}^{-1} \times 8 \mathrm{~h}}{2.303}$

$\Rightarrow \frac{[R]_{0}}{[R]}=\operatorname{antilog}(0.8024)$

$\Rightarrow \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}=6.3445$

$\Rightarrow \frac{[\mathrm{R}]}{[\mathrm{R}]_{0}}=0.1576$ (approx)

$ =0.158 $

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158 .

Answer

The given equation is

$k=\left(4.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{-28000 \mathrm{KT}(\mathrm{i})}$

Arrhenius equation is given by,

$k=\mathrm{Ae}^{-E_{\alpha} / \mathrm{R} T}$

From equation (i) and (ii), we obtain

$\frac{E_{a}}{\mathrm{R} T}=\frac{28000 \mathrm{~K}}{T}$

$\Rightarrow E_{a}=\mathrm{R} \times 28000 \mathrm{~K}$

$=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 28000 \mathrm{~K}$

$=232792 \mathrm{~J} \mathrm{~mol}^{-1}$

$=232.792 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Answer

Arrhenius equation is given by,

$k=\mathrm{Ae}^{-E_{\alpha} / \mathrm{R} T}$

$\Rightarrow \ln k=\ln \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$

$\Rightarrow \ln k=\log \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$

$\Rightarrow \log k=\log \mathrm{A}-\frac{E_{a}}{2.303 \mathrm{R} T} \quad \quad \quad \text{(i)}$

The given equation is

$\log k=14.34-1.25 \times 10^{4} \mathrm{~K} / T\quad \quad \quad \text{(ii)}$

From equation (i) and (ii), we obtain

$\frac{E_{a}}{2.303 \mathrm{R} T}=\frac{1.25 \times 10^{4} \mathrm{~K}}{T}$

$\Rightarrow E_{a}=1.25 \times 10^{4} \mathrm{~K} \times 2.303 \times \mathrm{R}$

$=1.25 \times 10^{4} \mathrm{~K} \times 2.303 \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

$=239339.3 \mathrm{~J} \mathrm{~mol} \cdot 1$ (approximately)

$=239.34 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Also, when $t_{1 / 2}=256$ minutes,

$ \begin{aligned} k & =\frac{0.693}{t_{1 / 2}} \\ & =\frac{0.693}{256} \end{aligned} $

$=2.707 \times 10^{-3} \mathrm{~min}^{-1}$

$=4.51 \times 10^{-5} \mathrm{~S}^{-1}$

It is also given that, $\log k=14.34-1.25 \times 10^{4} \mathrm{~K} / T$

$\Rightarrow \log \left(4.51 \times 10^{-5}\right)=14.34-\frac{1.25 \times 10^{4} \mathrm{~K}}{T}$

$\Rightarrow \log (0.654-05)=14.34-\frac{1.25 \times 10^{4} \mathrm{~K}}{T}$

$\Rightarrow \frac{1.25 \times 10^{4} \mathrm{~K}}{T}=18.686$

$\Rightarrow T=\frac{1.25 \times 10^{4} \mathrm{~K}}{18.686}$

$=668.95 \mathrm{~K}$

$=669 \mathrm{~K}$ (approximately)

Answer

From Arrhenius equation, we obtain

$\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$

Also, $k_{1}=4.5 \times 10^{3} \mathrm{~s}^{-1}$

$T_{1}=273+10=283 \mathrm{~K}$

$k_{2}=1.5 \times 10^{4} \mathrm{~s}^{-1}$

$E_{\mathrm{a}}=60 \mathrm{~kJ} \mathrm{~mol}^{-1}=6.0 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}$

Then,

$\log \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}}=\frac{6.0 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}\left(\frac{T_{2}-283}{283 T_{2}}\right)$

$\Rightarrow 0.5229=3133.627\left(\frac{T_{2}-283}{283 T_{2}}\right)$

$\Rightarrow \frac{0.5229 \times 283 T_{2}}{3133.627}=T_{2}-283$

$\Rightarrow 0.0472 T_{2}=T_{2}-283$

$\Rightarrow 0.9528 T_{2}=283$

$\Rightarrow T_{2}=297.019 \mathrm{~K}$ (approximately)

$=297 \mathrm{~K}$ $=24^{\circ} \mathrm{C}$

Hence, $k$ would be $1.5 \times 10^{4} \mathrm{~s}^{-1}$ at $24^{\circ} \mathrm{C}$.

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

Answer

For a first order reaction,

$t=\frac{2.303}{k} \log \frac{a}{a-x}$

At $298 \mathrm{~K}, \quad t=\frac{2.303}{k} \log \frac{100}{90}$

$=\frac{0.1054}{k}$

At $308 \mathrm{~K}, \quad t^{\prime}=\frac{2.303}{k^{\prime}} \log \frac{100}{75}$

$=\frac{2.2877}{k^{\prime}}$

According to the question,

$t=t^{\prime}$

$\Rightarrow \frac{0.1054}{k}=\frac{0.2877}{k^{\prime}}$

$\Rightarrow \frac{k^{\prime}}{k}=2.7296$

From Arrhenius equation, we obtain

$$ \begin{aligned} & \log \frac{k^{\prime}}{k}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{T^{\prime}-T}{T T^{\prime}}\right) \\ & \log (2.7296)=\frac{E_{a}}{2.303 \times 8.314}\left(\frac{308-298}{298 \times 308}\right) \\ & E_{a}=\frac{2.303 \times 8.314 \times 298 \times 308 \times \log (2.7296)}{308-298} \\ & \quad=76640.096 \mathrm{~J} \mathrm{~mol}^{-1} \\ & \quad=76.64 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$

To calculate $k$ at $318 \mathrm{~K}$,

It is given that, $A=4 \times 10^{10} \mathrm{~s}^{-1}, T=318 \mathrm{~K}$

Again, from Arrhenius equation, we obtain

$$ \begin{aligned} \log k & =\log A-\frac{E_{a}}{2.303 \mathrm{R} T} \\ & =\log \left(4 \times 10^{10}\right)-\frac{76.64 \times 10^{3}}{2.303 \times 8.314 \times 318} \\ & =(0.6021+10)-12.5876 \\ & =-1.9855 \end{aligned} $$

Therefore, $k=\operatorname{Antilog}(-1.9855)$

$$ =1.034 \times 10^{-2} \mathrm{~s}^{-1} $$

Answer

From Arrhenius equation, we obtain $\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$

It is given that, $k_{2}=4 k_{1}$

$T_{1}=293 \mathrm{~K}$

$T_{2}=313 \mathrm{~K}$

Therefore, $\log \frac{4 k_{1}}{k_{2}}=\frac{E_{a}}{2.303 \times 8.314}\left(\frac{313-293}{293 \times 313}\right)$

$\Rightarrow 0.6021=\frac{20 \times E_{a}}{2.303 \times 8.314 \times 293 \times 313}$

$\Rightarrow E_{\alpha}=\frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20}$

$$ =52863.33 \mathrm{~J} \mathrm{~mol}^{-1} $$

$$ =52.86 \mathrm{~kJ} \mathrm{~mol}^{-1} $$

Hence, the required energy of activation is $52.86 \mathrm{kJmol}^{-1}$.