# The activation energy of a certain reaction is 35.3 kJ/mol. At 20 degrees C, the rate constant is 0.0130 s^-1. At what temperature would this reaction go twice as fast? please help?

May 8, 2016

${T}_{2} = 308 K = {35}^{\circ} C$

#### Explanation:

To solve this question, we can modify the Arrhenius equation which provides a relationship between the equilibrium constant $k$ and temperature $T$:

$\ln k = - \frac{{E}_{a}}{R} \left(\frac{1}{T}\right) + \ln A$

To find the new temperature ${T}_{2}$ of the reaction in this question, we can use the following expression:

$\ln \left(\frac{{k}_{2}}{{k}_{1}}\right) = \frac{{E}_{a}}{R} \left(\frac{1}{{T}_{1}} - \frac{1}{{T}_{2}}\right)$

If the rate is doubled, the equilibrium constant will also double and therefore, ${k}_{2} = 2 \times {k}_{1} = 2 \times 0.0130 {s}^{- 1} = 0.0260 {s}^{- 1}$

Therefore, $\ln \left(\frac{0.0260 \cancel{{s}^{- 1}}}{0.0130 \cancel{{s}^{- 1}}}\right) = \frac{35300 \frac{\cancel{J}}{\cancel{m o l} \cdot \cancel{K}}}{8.31 \frac{\cancel{J}}{\cancel{m o l} \cdot \cancel{K}}} \left(\frac{1}{293 K} - \frac{1}{{T}_{2}}\right)$

Solving for ${T}_{2}$: $\implies {T}_{2} = 308 K = {35}^{\circ} C$

Here is a video that explains the theory behind this topic:
Chemical Kinetics | A Model for Chemical Kinetics & Catalysis.