Question

The rate of a reaction a doubles on increasing the temperature from 300 to 310 k. By how much, the temperature of reaction b should be increased from 300 k so that rate doubles if activation energy of the reaction b is twice to that of reaction

a.

Answer 1

Hey dear,

● Answer- 304.92 K

● Explanation-
# Given-
Ta1 = 300 K
Ta2 = 310 K
Tb1 = 300 K
Tb2 = x
Eb = 2 Ea

# Solution-
We have the relation,
Ea × (1/Ta1-1/Ta2) = Eb × (1/Tb1-1/Tb2)
Ea × (1/300-1/310) = 2Ea × (1/300-1/x)
(1/300-1/x) = 1/18600
1/x = 1/300 - 1/18600
1/x = 61/18600
x = 304.92 K

Temperature of reaction B need to be 304.92 K to double its rate.

Hope this is helpful...