Question

The initial concentration of the first order reaction:

N2O5 (g) → 2NO2 (g) + 1/2 O2 (g)

was 1.24 × 10⁻² mol L⁻¹ at 300 K. The concentration of N2O5 after 60 minutes was 0.20 × 10⁻² mol L⁻¹.

Calculate the rate constant of the reaction at 300 K.

Hint:

The units of the rate constant for a first-order reaction are always \( \text{time}^{-1} \). Ensure your concentrations are in the same units so they cancel out inside the log term.

Answer 1

Concept:
For a first-order reaction, the rate constant \( k \) is determined by the integrated rate law: $$ k = \frac{2.303}{t} \log \left(\frac{[R]_0}{[R]_t}\right) $$ Where:
[R]₀ = Initial concentration of the reactant.
[R]_t = Concentration of the reactant at time t.
t = Time elapsed.

Step 1: Identify the given values.
Initial concentration \( [R]_0 = 1.24 \times 10^{-2} \, \text{mol L}^{-1} \)
Final concentration \( [R]_t = 0.20 \times 10^{-2} \, \text{mol L}^{-1} \)
Time \( t = 60 \, \text{minutes} \)

Step 2: Apply the first-order rate equation.
$$ k = \frac{2.303}{60} \log \left(\frac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}}\right) $$ $$ k = \frac{2.303}{60} \log (6.2) $$
Step 3: Solve for \( k \).
Using \( \log(6.2) \approx 0.7924 \): $$ k = \frac{2.303 \times 0.7924}{60} $$ $$ k = \frac{1.8249}{60} $$ $$ k = 0.0304 \, \text{min}^{-1} $$