Chemical Kinetics: Application Problems (9)

For first order, $k = \ln 2/t_{1/2}$. At $300$ K:

Compute $1/T_1 - 1/T_2 = 1/300 - 1/320 = (320 - 300)/(300 \times 320) = 20/96000 = 2.083 \times 10^{-4}$ K$^{-1}$.

So $k_2 \approx 0.0347 \times 3.50 = 0.121$ min$^{-1}$.

Final answer: half-life $\approx 5.7$ min at $320$ K.

Why This Works

The Arrhenius equation $k = A e^{-E_a/RT}$ shows how rate constants depend on temperature. The ratio form eliminates the unknown $A$:

For first-order reactions, half-life is inversely proportional to $k$. So an increase in $k$ by factor 3.5 means the half-life decreases by factor 3.5.

The “rule of thumb” that reaction rates roughly double per $10$ K rise — this problem matches it well: $20$ K rise gave a ~3.5× speed-up, close to $2 \times 2 = 4$.

Alternative Method

Use logarithms in base 10:

This is the form CBSE uses. Same answer.