For following reactions A xrightarrow700 K Product A xrightarrow[ textcatalyst]500 K Product it was found that the Ea is decreased by 30 k J/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

Question

For following reactions A xrightarrow700 K Product A xrightarrow[ textcatalyst]500 K Product it was found that the Ea is decreased by 30 k J/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same): (A) 198 kJ/ mol (B) 105 kJ/ mol (C) 75 kJ/ mol (D) 135 kJ/ mol

Tardigrade Admin · Accepted Answer

K1=Ae (E a/R×700) K2=A× e((Ea-30)/R×500) For same rate K1=K2 e(E a/700 R)=e ((E a-30)/R×500) (E a/700 R)=(E a-30/R×500) 5 Ea=7Ea-210 210=2E a Ea=105 kJ/mole Ea-30=75

Q. For following reactions
$A 700 K$ Product
$A 500 K catalyst$ Product
it was found that the $E_{a}$ is decreased by $30 k J / m o l$ in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

$198 k J / m o l$

$105 k J / m o l$

$75 k J / m o l$

$135 k J / m o l$

Q. For following reactions A700K​ Product A500Kcatalyst​ Product it was found that the Ea​ is decreased by 30kJ/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

198kJ/mol

105kJ/mol

75kJ/mol

135kJ/mol

K1​=AeR×700Ea​ K2​=A×eR×500(Ea−30)​ For same rate K1​=K2​ e700REa​=e​R×500(Ea−30)​ 700REa​=R×500Ea−30​ 5Ea=7Ea−210 210=2Ea Ea​=105kJ/mole Ea​−30=75

Q. For following reactions
$A 700 K$ Product
$A 500 K catalyst$ Product
it was found that the $E_{a}$ is decreased by $30 k J / m o l$ in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

$198 k J / m o l$

$105 k J / m o l$

$75 k J / m o l$

$135 k J / m o l$