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

Answer

B

Explanation

In `(k_(2))/(k_(1))=(epsi_(o))/(R ) ((1)/(T_(1))-(1)/(T_(2)))`
In `2=(epsi_(o))/(8.314) (0.000107526) rArr epsi_(o) =53.47" kJ/mol"`
`epsi_(o)` for reaction B = 106.941 kJ/mol For reaction B
In `(k_(2))/(k_(1)) =(epsi_(o))/(R ) ((1)/(T_(1))-(1)/(T_(2)))`
In `2=(106941)/(8.314) ((1)/(300) -(1)/(T_(2))) rArr (0.693xx8.314)/(106941) =((1)/(300) -(1) /(T_(2)))`
`(1)/(T_(2)) =(1)/(300) -(0.6938.314)/(106941) rArr T_(2) =(1)/(0.003279) =304.9`
So, temperature should increased by 4.9 k.