Question

A reaction between substances Y and Z is

representedstochiometrically by
Y2 + Z2 ---> 2YZ


The rate constant obeys the Arrhenius equation. At 435.
K
the rate constant is k = 2.95 e-03 L/mol-s and A = 3.00 e+10
L/mol-s
What is the activation energy (KJ/mol) for
thisreaction?

Answer 1

Answer : The value of activation energy for this reaction is 108.318 kJ/mol

Explanation :

The Arrhenius equation is written as:

`K=A* e^{(-Ea)/(RT)}`

Taking logarithm on both the sides, we get:

`\ln k=-(Ea)/(RT)+\ln A` ............(1)

where,

k = rate constant =
`2.95* 10^(-3)L/mol.s`

Ea = activation energy = ?

T = temperature = 435 K

R = gas constant = 8.314 J/K.mole

A = pre-exponential factor =
`3.00* 10^(+10)L/mol.s`

Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:

`\ln (2.95* 10^(-3)L/mol.s)=-(Ea)/(8.314J/K.mol* 435K)+\ln (3.00* 10^(10)L/mol.s)`

`Ea=108318.365J/mol=108.318kJ/mol`

Therefore, the value of activation energy for this reaction is 108.318 kJ/mol