We’re being asked to **determine the temperature** at which the reaction would go twice as fast.

We’re given the rate constant at another temperature and the activation energy of the reaction. This means we need to use the ** two-point form of the Arrhenius Equation**:

$\overline{){\mathbf{ln}}{\mathbf{}}\frac{{\mathbf{k}}_{\mathbf{2}}}{{\mathbf{k}}_{\mathbf{1}}}{\mathbf{=}}{\mathbf{-}}\frac{{\mathbf{E}}_{\mathbf{a}}}{\mathbf{R}}{\mathbf{[}}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}{\mathbf{-}}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}{\mathbf{]}}}$

where:

**k _{1}** = rate constant at T

**k _{2}** = rate constant at T

**E _{a}** = activation energy (in J/mol)

**R** = gas constant (8.314 J/mol•K)

**T _{1} and T_{2}** = temperature (in K).

The activation energy of a certain reaction is 36.9 kJ/mol. At 20°C, the rate constant is 0.0140 s^{-1}. At what temperature in degrees Celsius would this reaction go twice as fast?

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