The correct option is B Temperature of the uncatalysed reaction is more than the catalysed reaction
Consider a rate of reaction with catalysed and uncatalysed condition.
k=Ae−Ea/RT......eqn(1)
kcat=Ae−E′a/RT.......eqn(2)
k : Rate constant without catalyst
kcat : Rate constant with catalyst
Ea : Activation energy without catalyst
E′a : Activation energy with catalyst
Assuming the rate of reaction of catalysed and the uncatalysed reactions are equal,
Ae−Ea/RT1=Ae−E′a/RT2
T1 : Temperature at which uncatalysed reaction occurs
T2 : Temperature at which catalysed reaction occur
e−Ea/RT1=e−E′a/RT2
Taking natural logarithm on both side
−EaRT1=−E′aRT2
EaT1=E′aT2
Since, Ea>E′a
T1>T2
Therefore, option (b) is correct.