Question

For the reaction \(2N_2O_5 \rightarrow 4NO_2 + O_2\), rate constant \(k = 4.48 \times 10^{-5} \, \text{s}^{-1}\) and the initial pressure is 600 atm. After 10 min, determine the final pressure of \(N_2O_5\):

• A. 590 atm
• B. 490 atm
• C. 588 atm
• D. 580 atm

Hint:

For first-order reactions, use the integrated rate law to calculate the final pressure, concentration, or volume after a given time.

Answer 1

The Correct Option is C

For a first-order reaction, the integrated rate law is: \[ \ln \frac{P_0}{P} = kt \] where: - \(P_0\) is the initial pressure, - \(P\) is the final pressure, - \(k\) is the rate constant, - \(t\) is time. Substitute the given values: \[ \ln \frac{600}{P} = 4.48 \times 10^{-5} \times 600 \] Solving for \(P\), we get 588 atm. Thus, the correct answer is Option (C).