Question

Many reactions double their rates with every ten degrees rise in temperature. Assume that such a reaction takes place at 301 K and 311 K. What must its activation energy be for this statement to hold

Answer 1

57.6 KJ/mol

Step-by-step explanation:

We have that;

T1 = 301 K

T2= 311 K

k1 = k1

k2= 2k1

R = 8.314 JK-1

Ea= ?

So;

ln(k1/k2) = (-Ea/R) (1/T1 - 1/T2)

ln (k1/2k1) = -(Ea/8.314) (1/301 - 1/311)

ln(1/2) = -(Ea/8.314) (3.3 * 10^-3 - 3.2 * 10^-3)

-0.693 = -(Ea/8.314) 0.1 * 10^-3

-0.693/ 0.1 * 10^-3 = -(Ea/8.314)

-6930 = -(Ea/8.314)

Ea = 6930 * 8.314

Ea = 57.6 KJ/mol