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

Answer:HEY BUDDY HERE IS YOUR ANSWER!!!

Explanation: Ta1 = 300 K

Ta 2 = 310 K

Tb 1 = 300 K

Tb 2 = 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.