Question

For a reaction rate constant is doubled when temperature is increased from 27°C to 37°C. What is the activation energy ?

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Answer 1

Explanation:

Given :

1. T1 = 27 + 273 = 300 K

2. T2 = 37 + 273 = 310 K

3. k1 = k and k2 = 2 k

Equation to find activation energy using temperature condition is given as :

⇒Log (k2 / k1) = Ea / 2.303 R [(T2 – T1) / T1 T2]

⇒Log (2 k / k) = Ea / 2.303 x 8.314 {(310 – 300) / 300 x 310}

⇒Log 2 = Ea / 2.303 x 8.314 x (10 / 300 x 310)

⇒Ea = 53598.6 J mol-1

⇒Ea = 53.6 kJ mol-1

Thus activation energy for the 10 degree rise in temperature is given as is 53.6 kJ mol-1

Answer 2

Answer:

Explanation:

We are given that:

When T1 = 27 + 273 = 300 K

Let k1 = k

When T2 = 37 + 273 = 310 K

k2 = 2 k

Substituting these values the equation:

Log (k2 / k1)

= Ea / 2.303 R {(T2 – T1) / T1 T2}

We will get:

Log (2 k / k) = Ea / 2.303 x 8.314 {(310 – 300) / 300 x 310}

Log 2 = Ea / 2.303 x 8.314 x (10 / 300 x 310)

Ea = 53598.6 J mol-1

Ea = 53.6 kJ mol-1

Hence, the energy of activation of the reaction is 53.6 kJ mol-1