Question

Two reactions of the same order have identical activation energies and their entropies of activations differ by 50 j k-1 mol-1. Calculate the ratio of their rate constants at any temperature.

Answer 1

to Arrhenius equation, K = Ae^{-Ea/RT}

Where A is proportionality constant, Ea is activation energy , R is universal has constant and T is temperature in kelvin.

Taking log both sides,

We get, look = logA - Ea/2.303RT

So, for 1st  reaction , logk₁ = logA - Ea₁/2.303RT

For 2nd reaction, logk₂ = logA - Ea₂/2.303RT

                       Subtracting equation (2) from equation (1),

logk₁ - logk₂ = (Ea₁ - Ea₂)/2.303RT

Here, (Ea₁ - Ea₂) = 24.9kJ

R = 8.3 J/K/mol

And T = 27°C = 300K

log(k₁/k₂) = (24.9 * 1000)/2.303 * 8.3 * 300 = 4.342

k₁/k₂ = antilog(4.342) = 2.198 × 10⁴

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