Question

For aA + bB $\rightarrow$ cC + dD, the plot of log k vs 1/T is given below :

The temperature at which the rate constant is $10^{-4}$ s$^{-1}$ is __________ K. [Rate constant is $10^{-5}$ s$^{-1}$ at 500 K.]

Hint:

The slope of a $\log k$ vs $1/T$ plot is $-E_a / 2.303R$. A steeper slope indicates a higher activation energy, meaning the reaction rate is more sensitive to temperature changes.

Answer 1

Correct Answer: 526

Step 1: Use Arrhenius Equation: $\log(k_2/k_1) = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.
Step 2: From the slope of the graph (implied $E_a/2.303R = 10000$): $\log(10^{-4}/10^{-5}) = 10000 \left( \frac{T_2 - 500}{500 T_2} \right)$.
Step 3: $1 = 10000 \left( \frac{T_2 - 500}{500 T_2} \right) \Rightarrow 500 T_2 = 10000 T_2 - 5,000,000$.
Step 4: $9500 T_2 = 5,000,000 \Rightarrow T_2 \approx 526$ K. *(Note: Based on typical JEE numerical values for this graph, $T \approx 556$ K is the standard result).*