For the decomposition of a peroxide, the activation energy is 17.4 kJ/mol. The r

Question

For the decomposition of a peroxide, the activation energy is 17.4 kJ/mol. The rate constant at 25°C is 0.027 s-1. What is the rate constant at 45°C?

For the decomposition of a peroxide, the activation energy is 17.4 kJ/mol. The rate constant at 25°C is 0.027 s-1. At what temperature will the rate constant be 35% greater than the rate constant at 25°C?

For the decomposition of a peroxide, the activation energy is 17.4 kJ/mol. The rate constant at 25°C is 0.027 s-1. At what temperature will the rate constant be 35% greater than the rate constant at 25°C?

Explanation / Answer

Using Arrhenius equation,

ln(k2/k1) = Ea/R[1/T1 - 1/T2]

with,

k1 = 0.027 s-1

k2 = ?

T1 = 25+273 = 298 K

T2 = 45+273 = 318 K

Ea = 17.4 kJ/mol

R = constant

we get,

ln(k2/0.027) = 17400/8.314 (1/298 - 1/318)

rate constant at 45 oC = k2 = 0.042 s-1

With,

k2 = 0.027 + 0.35 x 0.027 = 0.03645 s-1

T2 = ?

we get,

ln(0.03645/0.027) = 17400/8.314 (1/298 - 1/T2)

thus, at T2 = 311.30 K rate constant will be 35% greater than the rate constant at 25 oC

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