The reaction between ethyl iodide and hydroxide ion in ethanol (c2h5oh) solution

Question

The reaction between ethyl iodide and hydroxide ion in ethanol (c2h5oh) solution, c2h5oh + oh --> c2h5oh + I, has an activation energy of 86.8 kj/mol and a frequency factor of 2.1 x 10^ 11 and a rate constant, k, of 2.81 x 10^ -4 at 32 degrees Celcius

1) A solution of KOH in ethanol is made up by dissolving 0.341 g KOH in ethanol to form 289.0 ml of solution. Similarly, 1.523 g of C2H5I is dissolved in ethanol to form 289.0 of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at 32 degrees C?

2) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at 50 degrees C.

Explanation / Answer

well here we need to use the arrhenius equation which states that k=Ae^-Ea/rt so since the activation energy doesn't change, rate constant at 50 degrees=1.91*10^-3 s^-1 hope my answer is helpful

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