For a reaction \( xA \rightarrow yB, \, k = 0.3 \, \text{M}^{-1} \text{sec}^{-1} \),
(i) If the concentration of A is made 4 times, then the rate of reaction becomes 16 times.
(ii) The decomposition of \( \text{N}_2\text{O}_5 \) is an example of this type of reaction.
(iii) The order of the reaction is 2.
(iv) A plot of \( \ln \left( \frac{[A]_0}{[A]_t} \right) \) vs. \( t \) is a straight line.
(v) The half-life of the reaction is independent of the concentration of the reactant.
Which of the following options has the correct set of statements:
Show Hint
For second-order reactions, the rate is proportional to the square of the concentration, and the half-life depends on the initial concentration. A plot of \( \ln \left( \frac{[A]_0}{[A]_t} \right) \) vs. \( t \) for second-order reactions is linear.
Step 1: Identifying the type of reaction.
From the rate law \( k = 0.3 \, \text{M}^{-1} \text{sec}^{-1} \), the given rate constant indicates that the reaction follows second-order kinetics (rate proportional to the square of the concentration of \( A \)). Thus, the reaction is second-order.
Step 2: Analyzing the given statements.
- (i) The rate law for a second-order reaction is given by:
\[
\text{Rate} = k[A]^2
\]
If the concentration of A is quadrupled, the rate of reaction increases by \( 4^2 = 16 \), confirming statement (i).
- (ii) The decomposition of \( \text{N}_2\text{O}_5 \) follows second-order kinetics, confirming statement (ii).
- (iii) The reaction is indeed second-order, as discussed above. Hence, statement (iii) is correct.
- (iv) For a second-order reaction, a plot of \( \ln \left( \frac{[A]_0}{[A]_t} \right) \) vs. \( t \) should be a straight line, confirming statement (iv).
- (v) For a second-order reaction, the half-life is inversely proportional to the initial concentration of the reactant. Therefore, statement (v) is incorrect.
Step 3: Conclusion.
Thus, the correct set of statements is (i), (iii), and (v). Therefore, the correct answer is option (D).
Final Answer: (D) (i), (iii), (v)
