Grade 12Physical Chemistry
Please solve question no 29 that is question on activation energy
1 Answer
Vikas TU
From the activation energy formulae:
AT temperature 400 K, the time constant is with Ea activation energy,
K = Z*e^(-Ea/400R)............(1)
and
after decreasing we get,
AT temperature 300 K, the time constant is with Ea activation energy,
k = ze^(-0.05xEa/300R)..................(2)
Solving both the eqns. we get,
0.05x = 0.75
or
x = 15.
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