Question

For the reaction \( 2 N_2O_5 \rightarrow 4 NO_2 + O_2 \). If initial pressure is 100 atm and rate constant \( k \) is \( 3.38 \times 10^{-5} \, \text{s}^{-1} \), After 20 min the final pressure of \( N_2O_5 \) will be:

• A. 96 atm
• B. 50 atm
• C. 70 atm
• D. 60 atm

Hint:

For first-order reactions, the pressure decrease is logarithmic, and you can use the integrated rate law to calculate final pressures.

Answer 1

The Correct Option is A

Since the reaction is a first-order reaction, the integrated rate law for first-order reactions is given by: \[ \ln \left( \frac{P_0}{P_t} \right) = k \cdot t \] Where: - \( P_0 \) is the initial pressure = 100 atm - \( P_t \) is the pressure at time \( t \) - \( k \) is the rate constant = \( 3.38 \times 10^{-5} \, \text{s}^{-1} \) - \( t \) is the time = 20 minutes = 1200 seconds Substituting the values: \[ \ln \left( \frac{100}{P_t} \right) = (3.38 \times 10^{-5}) \times 1200 \] \[ \ln \left( \frac{100}{P_t} \right) = 0.04056 \] \[ \frac{100}{P_t} = e^{0.04056} \approx 1.0415 \] \[ P_t = \frac{100}{1.0415} = 96 \, \text{atm} \] Thus, the final pressure is 96 atm.