For first order decomposition of \( N_2O_5 \) at constant volume, \( p_i \) is initial pressure and \( p_t \) is pressure at any time \( t \), during the reaction. Which of the following statement/s is incorrect?
(A) Half life does not depend on initial pressure
(B) Total pressure decreases with time
(C) rate constant will be the same if initial pressure is doubled.
(D) \( k=\frac{2.303}{t}\log(\frac{p_t}{p_i}) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (D) only
For a first order gas-phase reaction at constant volume, the rate constant is written with the partial pressure of the reactant:
\( \displaystyle k = \frac{2.303}{t}\log\frac{p_{\ce{N2O5},\,0}}{p_{\ce{N2O5}}} \)
The decomposition gives more gas molecules than it uses:
\( \displaystyle \ce{2N2O5(g) -> 4NO2(g) + O2(g)} \)
If the partial pressure of \( \ce{N2O5} \) falls by 2p, the pressure of \( \ce{NO2} \) rises by 4p and that of \( \ce{O2} \) by p, so the total pressure is \( p_t = p_i + 3p \): the total pressure rises as the reaction proceeds.
(A) Half life does not depend on initial pressure — Correct. For a first order reaction \( t_{1/2} = 0.693/k \), which contains no initial concentration or pressure.
(B) Total pressure decreases with time — Incorrect. Two molecules of \( \ce{N2O5} \) give five molecules of gas, so the total pressure increases with time.
(C) rate constant will be the same if initial pressure is doubled. — Correct. The rate constant depends on temperature, not on the starting amount of reactant.
(D) \( k=\frac{2.303}{t}\log(\frac{p_t}{p_i}) \) — Incorrect. The ratio inside the log must be the initial over the present pressure of the reactant; \( p_t \) is the total pressure, which includes the products, so \( p_t/p_i \) is not that ratio.