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, C2H5I(alc) + OH -(alc) C2H5OH(l) + I -(alc) has an activation energy of 86.8 kj/mol and a frequency factor of 2.1*10^11 A solution of KOH in ethanol is made up by dissolving 0.319 gKOH in ethanol to form 300.0mL of solution. Similarly, 1.552g of C2H5I is dissolved in ethanol to form 300.0mL 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 45C?

Explanation / Answer

k = A · e^( -Ea / ( R·T)) Here A = 2.1×10¹¹ M?¹s?¹ Ea = 86.8kJ/mol = 86800J/mol T = 35°C = 308.15K => k = 2.1×10¹¹ M?¹s?¹ · e^( - 86800J/mol / (8.314472J/molK · 308.15K)) = 4.06×10?4 M?¹s?¹

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