The rate constant for the reaction 1 doubles on raising the temperature from 410^$\circ$C to 425^$\circ$C. The activation energy, in J/mol, is

Question

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The Arrhenius equation relates the rate constant (k) to the activation energy (Ea), temperature (T), and a pre-exponential factor (A). It is given by:

k = A e^(-Ea/RT)

where R is the gas constant.

If we take the natural logarithm of both sides and rearrange, we get:

ln(k2/k1) = (-Ea/R) (1/T2 - 1/T1)

where k1 and k2 are the rate constants at temperatures T1 and T2, respectively.

In this problem, we are given that k2 = 2k1 (since the rate constant doubles), T1 = 410°C, and T2 = 425°C. We can then solve for Ea:

ln(2) = (-Ea/R) (1/698 K - 1/698.15 K)

Ea = -ln(2) * R / (1/698 K - 1/698.15 K) = 1.8 × 10^15 J/mol

We can then convert this to the units given in the answer choices (10^14 or 10^15) and choose the closest option, which is B (1.8 × 10^14 J/mol). Note that the negative sign in option E indicates an invalid answer (activation energy must be positive).

The Rate Constant for the Reaction 1 Doubles on Raising