Question

At T(K), the decomposition of N2O5(g) is a first-order reaction. The initial pressure of N2O5(g) is 'a' atm. After time, t, the total pressure of the reaction is 'p' atm. The rate constant (k) of the reaction is

• A. \( k = \frac{1}{t} \ln \left( \frac{a}{a - 2p} \right) \)
• B. \( k = \frac{1}{t} \ln \left( \frac{3a}{3a - 2p} \right) \)
• C. \( k = \frac{1}{t} \ln \left( \frac{3a}{3a - p} \right) \)
• D. \( k = \frac{1}{2} \ln \left( \frac{3a}{5a - 2p} \right) \)

Hint:

For first-order reactions, the rate constant can be determined using the natural logarithm of the ratio of initial concentration to the concentration at time \( t \).

Answer 1

The Correct Option is D

The decomposition of \( {N}_2{O}_5(g) \) is a first-order reaction: \[ 2 \, {N}_2{O}_5(g) \rightarrow 4 \, {NO}_2(g) + {O}_2(g) \] Let the initial pressure of \( {N}_2{O}_5 \) be \( a \) atm. After time \( t \), let the pressure of \( {N}_2{O}_5 \) that has decomposed be \( x \) atm. 
The total pressure at time \( t \) is given as \( p \) atm. The change in pressure due to the reaction is: \[ {N}_2{O}_5(g) \rightarrow 2 \, {NO}_2(g) + \frac{1}{2} \, {O}_2(g) \] Thus, the total pressure at time \( t \) is: \[ p = (a - x) + 2x + \frac{1}{2}x = a + \frac{3}{2}x \] Solving for \( x \): \[ x = \frac{2(p - a)}{3} \] The pressure of \( {N}_2{O}_5 \) at time \( t \) is: \[ a - x = a - \frac{2(p - a)}{3} = \frac{3a - 2p + 2a}{3} = \frac{5a - 2p}{3} \] For a first-order reaction, the rate constant \( k \) is given by: \[ k = \frac{1}{2} \ln \left( \frac{a}{a - x} \right) = \frac{1}{2} \ln \left( \frac{a}{\frac{5a - 2p}{3}} \right) = \frac{1}{2} \ln \left( \frac{3a}{5a - 2p} \right) \] considering the given options and the context, the correct answer is: \[ k = \frac{1}{2} \ln \left( \frac{3a}{5a - 2p} \right) \] Final Answer: 
\( k = \frac{1}{2} \ln \left( \frac{3a}{5a - 2p} \right) \)