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

Options

A. 9.84K
B. 4.92 K
C. 2.45K
D. 19.67K

Answer

B

Explanation

`log ""(K_(2))/(K_(1))=(E)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
For reaction A -
Given, `(K_(2))/(K_(1))= 2, T_(1)=300 K, T_(2) = 310 K`
`log2 = (E)/(2.303R)[(1)/(300)-(1)/(310)]" "(i)`
For reaction B-
Given, `(K_(2))/(K_(1))=2, E = 2E, T_(1)=300 K, T_(2)=?`
`log 2=(E)/(2.303 R)[(1)/(300)-(1)/(T_(2))]" "(ii)`
From equation (i) and (ii),
`(2E)/(2.303R)[(1)/(300)-(1)/(T_(2))]=(E)/(2.303R)[(1)/(300)-(1)/(310)]`
`=2[(1)/(300)-(1)/(T_(2))]=(310 - 300)/(300 xx 310)`
`impliesT_(2)=304.92 K`
`T_(1)=300 K, T_(2)=304.92 K`
`Delta T=T_(2)-T_(1)=4.92 K`