aAt 34 C a reaction produces product at the rate of 0162 mol

a)At 34 °C, a reaction produces product at the rate of 0.162 mol L -1 h -1 . If the activation energy is 6 2 kJ/mol, what will the reaction rate be at 58 °C b) A reaction proceeds with H = -80 kJ/mol. The energy of activation of the uncatalyzed reaction is 80 kJ/mol, whereas it is 55 kJ/mol for the catalyzed reaction. How many times faster is the catalyzed reaction than the uncatalyzed reaction at 25°C? Express your answer in scientific notation to two significant figures

Solution

a) Use the Arrhenius equation as

ln k = ln A – E_a/RT

where k = rate constant for the reaction at temperature T and E_a is the activation energy for the reaction.

We have k = 0.162 mol/L.h at T = 34°C = (273 + 34) K = 307 K and E_a = 6.2 kJ/mol. We are required to find out the rate constant, k’ at T = 58°C = (273 + 58) K = 331 K.

Write down the integrated forms of the Arrhenius equation as

ln (0.162 mol/L.h) = ln A – (6.2 kJ/mol)/(8.314 J/mol.K)*1/(307 K) …..(1)

ln k’ = ln A – (6.2 kJ/mol)/(8.314 J/mol.K)*1/(331 K) ……(2)

Subtract (2) from (1) to get

ln (0.162 mol/L.h)/k’ = -(6.2 kJ/mol)/(8.314 J/mol.K)*(1/307 – 1/331) K^-1

=====> ln (0.162 mol/L.h)/k’ = (6.2 kJ/mol)*(1000 J/1 kJ)/(8.314 J/mol)*(0.003257 – 0.003021)

=====> ln (0.162 mol/L.h)/k’ = -(745.730)*(2.36*10^-4) = -0.17599

=====> (0.162 mol/L.h)/k’ = exp(-0.17599) = 0.83863

=====> k’ = (0.162 mol/L.h)/(0.83863) = 0.193 mol/L.h (ans).

b) As before, we have

k₁ = A*exp(-E_a1/RT) …….(1)

k₂ = A*exp(-E_a2/RT) …….(2)

where k₁ and k₂ are the rate constants for the uncatalyzed and catalyzed reactions.

Divide (1) by (2) to get

k₁/k₂ = exp[-1/RT*(E_a1 – E_a2)]

We have T = 25°C = (25 + 273) K = 298 K; E_a1 = 80 kJ/mol and E_a2 = 55 kJ/mol. Plug in values and obtain

k₁/k₂ = exp[-1/(8.314 J/mol.K)(298 K)*(80 kJ/mol – 55 kJ/mol)] = exp[-1/(2477.572 J/mol)*(25 kJ/mol)] = exp[-1/(2477.572 J/mol)*(25 kJ/mol)*(1000 J/1 kJ)] = exp[-1/(2477.572 J/mol)*(25000 J/mol)] = exp(-10.09052) = 4.14708*10^-5

Therefore, k₂/k₁ = 1/(4.14708*10^-5) = 24113.35205 = 2.411335205*10⁴ 2.4*10⁴

The catalyzed reaction is 2.4*10⁴ times faster than the uncatalyzed reaction at 25°C (ans).