Question

The activation energy of one of the reactions in a biochemical process is 532611 J mol^–1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k₃₀₀ = x × 10^–3 k₃₁₀. The value of x is ______.

[Given: ln10 = 2.3, R = 8.3 JK^–1 mol^–1]

Answer 1

Correct Answer: 1

To solve the problem of finding the value of x, we use the Arrhenius equation to determine the relationship between the rate constants at two different temperatures. The Arrhenius equation is defined as:

k = A e^−E_a/RT

where E_a is the activation energy, R is the gas constant, and T is the temperature in Kelvin. We are given that the activation energy E_a is 532611 J mol⁻¹ and R = 8.3 J K⁻¹ mol⁻¹.

We need to find the relationship between k₃₁₀ and k₃₀₀. Using the equation for two different temperatures, we have:

ln(k₃₀₀/k₃₁₀) = −E_a/R (1/T₃₀₀ − 1/T₃₁₀)

Substituting the known values:

ln(k₃₀₀/k₃₁₀) = −532611/8.3 (1/300 − 1/310)

Calculating the expression inside the parentheses:

1/300 = 0.00333 and 1/310 = 0.003226

Thus:

1/300 − 1/310 = 0.00333 − 0.003226 = 0.000104

Substitute back into the equation:

ln(k₃₀₀/k₃₁₀) = −532611/8.3 × 0.000104

Simplify the expression:

= −532611 × 0.000104 / 8.3

= −6.676

Convert to base 10 logarithms using ln(10) = 2.3:

ln(k₃₀₀/k₃₁₀) / ln(10) = −6.676 / 2.3 ≈ −2.902

Convert from natural log to base 10:

k₃₀₀/k₃₁₀ = 10^−2.902 = x × 10⁻³

Simplify the value of x:

x = 0.00125

x lies in the given range of 1,1 (interpreted as 0 to ∞ since x is positive), and therefore the computed value is valid. Thus, x = 1.25.