Question

A certain reaction has an activation energy of 50.84 kJ/mol. At what Kelvin temperature will the reaction proceed 5.00 times faster than it did at 319 K?

what is the value in K?

Answer 1

We know that

$Ea = 2.303 R (T1T2 / T2-T1) log K2 / K1$

Where Ea = Energy of activation = 50.84kJ/mol = 50840 J/mol (given in question)
R = molar gas constant = 8.314 J / ⁰C / mole
T1 = 319 K(given),

k1 = rate constant at T1
T2 = ? ,

k2 = rate constant at T2

As given in question

K2 = 5 X k1
on placing the values
50840 = 2.303x8.314x319xT2/(T2-319) x log 5k1/k1
50840 = 2.303x8.314x319xT2/(T2-319) x log 5
50840= 6107.9 T2/(T2-319) x 0.6989 = 4268.81 T2/(T2-319)
so 50840x(T2-319) = 4268.81T2
or 50840 T2 - 16217960 = 4268.81T2

Bringing T2 terms on one side
or (50840 - 4268.81) T2 = 16217960
or 46571.19 T2 = 16217960
or T2 = 348.24 K