We’re being asked to **determine the k at 75°C**

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**:

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

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 rate constant of a reaction is 4.7 x 10^{-3} s^{-1} at 25°C, and the activation energy is 33.6 kJ/mol. What is *k* at 75°C?

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.