A reaction has a rate constant of 1.23 Solution A.) Note that ln(a) - ln(b) = ln

Question

A reaction has a rate constant of 1.23

Explanation / Answer

A.) Note that ln(a) - ln(b) = ln(a/b) Solve you equation for the activation energy Ea: ln(k2) = ln(k1) + (Ea/R)·(1/T1 - 1/T2) ln(k2 /k1) = (Ea/R)*(1/T1 - 1/T2) Ea = R·ln(k2 /k1)/(1/T1 - 1/T2) R is the universal gas constant, k2 and k1 are the rate constant at absolute temperatures T2 and T1. It plays no role which rate constant you label as k2 and k1. There result is the same. So set k1 = 1.23×10^-4 s^-1 at T1 = (28 + 273)K = 301K k2 = 0.234s?¹ at T2 = (79 + 273)K = 352K => Ea = 8.3145Jmol?¹K?¹ · ln(0.234s?¹ / 1.23×10?4s?¹) / (1/301K - 1/352K) = 130,429Jmol?¹ = 130.429 kJmol?¹ B. Let k2 be the rate constant you're looking for at a temperature of T2 = (15 + 273)K = 288K ln(k2) = ln(k1) + (Ea/R)·(1/T1 - 1/T2) => e^(ln(k2)) = e^( ln(k1) + (Ea/R)·(1/T1 - 1/T2) ) k2 = e^( ln(k1) )·e^( (Ea/R)·(1/T1 - 1/T2) ) k2 = k1·e^( (Ea/R)·(1/T1 - 1/T2) ) It makes no difference which of the two known rate constants you choose to be k1 at T1. Let's pick the first one: k2 = 1.23×10^-4 s?¹ · e^( (130,429Jmol?¹ / 8.3145Jmol?¹K?¹)·(1/301K - 1/288K) ) = 1.23×10^-4 s?¹ · e^( -2.3524 ) = 0.0000117023 s?¹

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