Question

For a particular reaction variation of rate constant with temperature is given by

ln k_t = ln k₀ + $\left( \frac{\mathcal{l}n3}{10} \right)t$ (t ≥ 0°C)

Temp. Coefficient of reaction is –

• A. 4
• B. 3
• C. 10
• D. 2

Answer 1

Answer: (B)

3

Answer 2

We know $k_{t_{2}}$= $k_{t_{1}}$ (m) $\frac{t_{2}–t_{1}}{10}$

OR ln $k_{t_{2}}$ = ln $k_{t_{1}}$+ $\frac{t_{2}–t_{1}}{10}$ ln m

putting t₁ = 0

ln $k_{t_{2}}$ = ln k + $\frac{t_{2}}{10}$ ln m

Comparing with ln k_t = ln k₀ + $\frac{\mathcal{l}n3}{3}t$

M = 3