We need to use the ** two-point form of the Arrhenius Equation**:

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

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).

A certain reaction has an activation energy of 37.51 kJ/mol. At what Kelvin temperature will the reaction proceed 5.00 times faster than it did at 349 K?

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