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Chemical Kinetics — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryChemical Kinetics4 Marks
Question
The following reaction, $\text{A}_{(\text{g})}\xrightarrow{\ \ \triangle\ \ \ }\text{P}_{(\text{g})}+\text{Q}_{(\text{g})}+\text{R}_{(\text{g})},$ follows first order kinetics. The half-life period of this reaction is $69.3s$ at $500^\circ C$. The gas A is enclosed in a container at $500^\circ C$ and at a pressure of $0.4$ atm.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The rate constant for the reaction is:
  1. $0.4s^{-1}$
  2. $0.02s^{-1}$
  3. $0.01s^{-1}$
  4. $0.3s^{-1}$
  1. The pressure of the gas $A$ after $230$ s will be:
  1. $0.04$ atm
  2. $0.36$ atm
  3. $0.4$ atm
  4. $0.036$ atm
  1. The total pressure of the system after $230$ swill be:
  1. $2.15$ atm
  2. $1.12$ atm
  3. $0.4$ atm
  4. $3.08$ atm
  1. The plot ofln[A] vs twill be:
  1. Linear with slope $= k$
  2. Linear with intercept $= In[A]_0$
  3. Linear with slope $= In[A]_0$
  4. Linear with intercept $= [A]_0$
  1. Which of the following is not an example of first order reaction?
  1. $\text{C}_2\text{H}_{4(\text{g})}+\text{H}_{2(\text{g})}\rightarrow\text{C}_2\text{H}_{6(\text{g})}$
  2. $2\text{N}_2\text{O}_{5(\text{g})}\rightarrow4\text{NO}_{2(\text{g})}+\text{O}_{2(\text{g})}$
  3. $2\text{N}\text{H}_{3(\text{g})}\xrightarrow[\triangle]{\text{pt}}\text{N}_{2(\text{g})}+3\text{H}_{2(\text{g})}$
  4. $2\text{N}_2\text{O}_{(\text{g})}\xrightarrow{\ \ \triangle\ \ }2\text{N}_{2(\text{g})}+\text{O}_{2(\text{g})}$
✓

Answer

  1. (c) $0.01s^{-1}$
Explanation:
$\text{t}_\frac{1}{2}=69.3\text{s}$
For first order reaction,
$\text{k}=\frac{0.693}{\text{t}_\frac{1}{2}}=\frac{0.693}{69.3}=0.01\text{s}^{-1}$
  1. (a) $0.04$ atm
Explanation:
$\text{k}=\frac{2.303}{\text{t}}\log\frac{\text{a}}{\text{a}-\text{x}}=\frac{2.303}{230}\log\frac{0.4}{0.4-\text{x}}$
$0.01=\frac{2.303}{230}\log\frac{0.4}{0.4-\text{x}}=0.01\log\frac{0.4}{0.4-\text{x}}$
$\frac{0.01}{0.01}=\log\frac{0.4}{0.4-\text{x}}\Rightarrow1=\log\frac{0.4}{0.4-\text{x}}$
Antilog of $1=\frac{0.4}{0.4-\text{x}}\Rightarrow10=\frac{0.4}{0.4-\text{x}}\Rightarrow\text{x}=0.36$
$\therefore\text{a}-\text{x}=0.4-0.36=0.04\text{ atm}$
  1. (b) $1.12$ atm
Explanation:
For the given reaction,
$\text{A}_{(\text{g})}\xrightarrow{\ \ \triangle\ \ \ }\text{P}_{(\text{g})}+\text{Q}_{(\text{g})}+\text{R}_{(\text{g})}$
Initial pressur
$0.4$
$0$
$0$
$0$
Final pressure
$0.4-0.36$
$0.36$
$0.36$
$0.36$
Total pressure $= (0.4 - 0.36) + (3 \times 0.36) = 1.12$ atm
  1. (b) Linear with intercept $= In[A]_0$
Explanation:
Expression that relates concentration of reactant and ti me for first order reaction is
In$[A] = -kt + In[A]_0$
So, the plot of In [A] vs twill be linear with slope $= -k$ and intercept $= In[A]_0$
  1. (c) $2\text{N}\text{H}_{3(\text{g})}\xrightarrow[\triangle]{\text{pt}}\text{N}_{2(\text{g})}+3\text{H}_{2(\text{g})}$
Explanation:
Decomposition of ammonia on a hot platinum surface at high pressure is a zero order reaction.

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