Question

0.0210 M solution of \( \mathrm{N_2O_5} \) is allowed to decompose at \( 43^\circ\mathrm{C} \). How long will it take to reduce to 0.0150 M? (Given \( k = 6.0 \times 10^{-4}\ \mathrm{s^{-1}} \))

• A. 5600 s
• B. 360.0 s
• C. 560.0 s
• D. 3364 s

Hint:

For first order reactions, concentration change depends logarithmically on time.

Answer 1

The Correct Option is C

Step 1: Identifying reaction order.
Decomposition of \( \mathrm{N_2O_5} \) follows first order kinetics.

Step 2: Using first order rate equation.
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \]

Step 3: Substituting given values.
\[ t = \frac{2.303}{6.0 \times 10^{-4}} \log \frac{0.0210}{0.0150} \]

Step 4: Calculation.
\[ t = \frac{2.303}{6.0 \times 10^{-4}} \times 0.1461 \approx 560\ \text{s} \]

Step 5: Conclusion.
Time required is 560 seconds.