Question

Consider the following reactions:

$A \longrightarrow X, \ k_A = 10^{18}e^{-8000/T}$

$B \longrightarrow X, \ k_B = 10^{17}e^{-4000/T}$

Find the temperature $T$ (in K) at which $k_A = k_B$.

• A. 4000 K
• B. 8000 K
• C. $\frac{4000}{2.303}$ K
• D. $\frac{8000}{2.303}$ K

Answer 1

Answer: (C)

$\frac{4000}{2.303}$ K

Answer 2

To find the temperature $T$ at which $k_A = k_B$, we set the two rate constants equal to each other:

$10^{18}e^{-8000/T} = 10^{17}e^{-4000/T}$

Dividing both sides by $10^{17}$ gives:

$10e^{-8000/T} = e^{-4000/T}$

Rearranging the terms:

$10 = \frac{e^{-4000/T}}{e^{-8000/T}} = e^{(-4000/T) - (-8000/T)} = e^{4000/T}$

Taking the natural logarithm of both sides:

$\ln(10) = \frac{4000}{T}$

Since $\ln(10) \approx 2.303$:

$2.303 = \frac{4000}{T}$

Solving for $T$:

$T = \frac{4000}{2.303}$ K