Question

For the reaction, following data is given, A → B; $K₁ = 10¹⁴ × e^(⁻¹⁵⁰⁰/T)$ C → D; $K₂ = 10¹² × e^(⁻¹⁰⁰⁰/T)$ The temperature at which $K₁ = K₂$ is:

• A. 118.55 K
• B. 108.55 K
• C. 468.4 K
• D. 434.22 K

Answer 1

Answer: (B)

108.55 K

Answer 2

We are given the rate constants for two reactions as: $K₁ = 10^{14} \times e^{-1500/T}$ $K₂ = 10^{12} \times e^{-1000/T}$

We need to find the temperature T at which $K₁ = K₂$. Set the two expressions for K equal to each other: $10^{14} \times e^{-1500/T} = 10^{12} \times e^{-1000/T}$

Divide both sides by $10^{12}$: $10^2 \times e^{-1500/T} = e^{-1000/T}$

Take the natural logarithm of both sides: $\ln(10^2) + \ln(e^{-1500/T}) = \ln(e^{-1000/T})$ $2 \ln(10) - \frac{1500}{T} = -\frac{1000}{T}$

Rearrange the terms to solve for T: $2 \ln(10) = \frac{1500}{T} - \frac{1000}{T}$ $2 \ln(10) = \frac{500}{T}$

Now, solve for T: $T = \frac{500}{2 \ln(10)}$

Using the approximate value $\ln(10) \approx 2.3026$: $T = \frac{500}{2 \times 2.3026} = \frac{500}{4.6052} \approx 108.57$ K

The closest option is 108.55 K.