The rate constants for the first-order decomposition of a compound are 5.22× 10^-4 s^-1 at 43°C and 2.91 × 10^-3 s^-1 at 62°C.What is the value of the activation energy for this reaction? (R = 8.31 J/(mol · K))

Question

Quizplus · Accepted Answer

We can use the Arrhenius equation to solve for the activation energy (Ea):

$k = Ae^{-Ea/RT}$

Taking the natural logarithm of both sides:

ln($k$) = ln($A$) - $\frac{Ea}{RT}$

We can set up two equations using the given rate constants and temperatures:

ln($k_1$) = ln($A$) - $\frac{Ea}{R(316)}$

ln($k_2$) = ln($A$) - $\frac{Ea}{R(335)}$

where $k_1$ and $k_2$ are the rate constants at 43°C and 62°C, respectively.

Solving for ln($A$) by subtracting the second equation from the first:

ln($k_1$) - ln($k_2$) = $\frac{Ea}{R}(\frac{1}{T_2} - \frac{1}{T_1})$

Plugging in the given values and solving for Ea:

ln(5.22x10^-14) - ln(2.91x10^-13) = $\frac{Ea}{8.31}(1/335 - 1/316)$

Ea = 79.5 kJ/mol

Therefore, the answer is D.

The Rate Constants for the First-Order Decomposition of a Compound