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 2.10×1011M1s1.

A.)Predict the rate constant for the reaction at 36 C.

B.)A solution of KOH in ethanol is made up by dissolving 0.341 g KOH in ethanol to form 251.0 mL of solution. Similarly, 1.451 g of C2H5I is dissolved in ethanol to form 251.0 mL 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 36 C?

Explanation / Answer

According to Arrhenius rate equation

k= A exp(-Ea/RT) where k is rate constant,Ea is activation energy, R is gas constant, T is temperature in K, A is pre-exponentail factor

Ea= 86.8 kJ/ mom =86800 J/ mol

T = 36oC =309K

R= 8.314 J/ K mol

A=2.1*1011M-1s-1

k=2.1*1011*exp(-86800/8.314*309) = 4.4531*10-4 M-1s-1

b) weight of C2H5I (w2) = 1.451g

Molecular mass of C2H5I = 155.9656 g/mol

Volume of solution prepared (V)= 251 mL

Molarity of C2H5I = w2*1000/ molecular mass* V = (1.451*1000)/(155.9656*251) = 0.03706 mol/L

For a first order reaction

Rate =k[C2H5I]

k we get from first question .we are mixing equal volume of two solutions. So the concentration of C2H5I decrdecrease by amount 2.( we are mixing 1 litre of both solution.so total volume will be 2litre. So molarity of C2H5I decrease by 2)

New molarity = 0.03706/2 = 0.01853 mol/ L.

Rate = 4.4531*10-4 *0.01853 = 8.2516*10-6 sec-1

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