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Chemical Kinetics — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryChemical Kinetics4 Marks
Question
For the reaction : $2\text{NO}_\text{(g)}+\text{Cl}_{2\text{(g)}}\rightarrow2\text{NOCl}_\text{(g)},$ the following data were collected. All the measurements were taken at $263K.$
Experiment No.
Initial [NO] (M)
Initial $[Cl_2]$ (M)
Initial rate of disapp. of $Cl_2 $ (M/ min)
$1.$
$0.15$
$0.15$
$0.60$
$2.$
$0.15$
$0.30$
$1.20$
$3.$
$0.30$
$0.15$
$2.40$
$4.$
$0.25$
$0.25$
$?$
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The molecularity of the reaction is:
  1. $1$
  2. $2$
  3. $3$
  4. $4$
  1. The expression for rate law is:
  1. $\text{r}=\text{k}[\text{NO}][\text{Cl}_2]$
  2. $\text{r}=\text{k}[\text{NO}]^2[\text{Cl}_2]$
  3. $\text{r}=\text{k}[\text{NO}][\text{Cl}_2]^2$
  4. $\text{r}=\text{k}[\text{NO}]^2[\text{Cl}_2]^2$
  1. The overall order of the reaction is:
  1. $2$
  2. $0$
  3. $1$
  4. $3$
  1. The value of rate constant is:
  1. $150.32\ M^{-2} \min^{-1}$
  2. $200.08\ M^{-1} \min^{-1}$
  3. $177.77\ M^{-2} \min^{-1}$
  4. $155.75\ M^{-1} \min^{-1}$
  1. The initial rate of disappearance of $Cl_2$ in experiment $4$ is:
  1. $1.75M\ \min^{-1}$
  2. $3.23M\ \min^{-1}$
  3. $2.25M\ \min^{-1}$
  4. $2.77M\ \min^{-1}$
✓

Answer

  1. (c) $3$
Explanation:
$2\text{NO}_\text{(g)}+\text{Cl}_{2\text{(g)}}\rightarrow2\text{NOCl}_\text{(g)}$
Molecularity $= 3$
  1. (b) $\text{r}=\text{k}[\text{NO}]^2[\text{Cl}_2]$
Explanation:
Let rate of this reaction, $\text{r}=\text{k}[\text{NO]}^\text{m}[\text{Cl}_2]^\text{n}$
then $\frac{\text{r}_1}{\text{r}_2}=\frac{0.60}{1.20}=\frac{\text{k}(0.15)^\text{m}(0.15)^\text{n}}{\text{k}(0.15)^\text{m}(0.30)^\text{n}}$
or, $\frac{1}{2}=\Big(\frac{1}{2}\Big)^\text{n}\Rightarrow\text{n}=1$
Again from $\frac{\text{r}_2}{\text{r}_3}=\frac{1.20}{2.40}=\frac{\text{k}(0.15)^\text{m}(0.30)^\text{n}}{\text{k}(0.30)^\text{m}(0.15)^\text{n}}$
or $\frac{1}{2}=\Big(\frac{1}{2}\Big)^\text{m}\cdot\frac{2}{1}$ or $\frac{1}{4}=\Big(\frac{1}{2}\Big)^\text{m}\Rightarrow\text{m}=2$
Hence, expression for rate law is
$\text{r}=\text{k}[\text{NO}]^2[\text{Cl}_2]^1$
  1. (d) $3$
Explanation:
As the order w.r.t. $NO$ is $2$ and order w.r.t. $Cl_2$ is $1,$ hence the overall order is $3.$
  1. (c) $177.77\ M^{-2}\ \min^{-1}$
Explanation:
Substituting the values of experiment 1 in rate law expression
$0.60M\ \min^{-1} = k(0.15M)^2 (0.15M)^1$
or $\text{k}=\frac{0.60\text{M min}^{-1}}{0.0225\times0.15\text{M}^3}=177.77\text{M}^{-2}\text{min}^{-1}$
  1. (d) $2.77  M\ \min^{-1}$​​​​​​​
Explanation:
$r = 177.7  M^{-2} \ \min^{-1} \times (0.25M)^2 (0.25M)$
$= 2.77  M \ \min^{-1}$

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