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Grade 11Engineering Entrance Exams

The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be: (R = 8.314 J/K/ mol and log 2 = 0.301) ?

Profile image of Jayant Kumar
12 Years agoGrade 11
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1 Answer

Profile image of Saurabh Koranglekar
6 Years ago

To determine the activation energy (Ea) of the reaction, we use the Arrhenius equation in its logarithmic form:

k₂/k₁ = e^[(Ea/R) * (1/T₁ - 1/T₂)]

Taking the natural logarithm on both sides:

ln(k₂/k₁) = (Ea/R) * (1/T₁ - 1/T₂)

Given that the rate doubles when temperature increases from 300 K to 310 K, we have:

k₂/k₁ = 2

Taking the logarithm (base 10) on both sides:

log 2 = (Ea / 2.303R) * (1/T₁ - 1/T₂)

Substituting the given values:

0.301 = (Ea / (2.303 * 8.314)) * (1/300 - 1/310)

First, calculate (1/300 - 1/310):

1/300 = 0.003333
1/310 = 0.003226

(1/300 - 1/310) = 0.000107

Now, calculate 2.303 * 8.314:

2.303 * 8.314 = 19.14

Now solve for Ea:

0.301 = (Ea / 19.14) * 0.000107

Ea = (0.301 * 19.14) / 0.000107

Ea = 53858.88 J/mol

Ea ≈ 53.86 kJ/mol

Final Answer: Activation energy (Ea) ≈ 53.86 kJ/mol


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