Question

A particular reaction has an activation energy, E_a, of 112 kJ/mol.  If the rate constant for the reaction is 0.00621 s ⁻¹ at 452 °C, at what temperature(in°C) would the rate constant be 0.145 s ⁻¹?

Answer 1

Answer -

Given,

Activation Energy = 112 kJ/mol

Rate constant (k1) at 452 C (T1) = 0.00621 s−1

Temperature (T2) at which Rate constant (k2) is 0.145 s−1 = ?

We know that,

k = A.e-(Ea/RT) -----------------------A

where,

k = Rate

A = frequency factor

Ea = Activation Energy

R = Gas Constant (8.3145 J/Kmol)

T = Temperature in K

We know that,

°C + 273.15 = K

So,

452 C + 273.15 = 725.15 K

Also,

1 kJ/mol = 1000 J/mol

So, 112 kJ/mol = 112 * 1000 J/mol

112 kJ/mol = 112000 J/mol

Put the initial values in A,

k = A.e-(Ea/RT)

0.00621 s−1 = A.e-(112000 J/mol) / [(8.3145 J/Kmol)* (725.15 K​​​​​​​)]

0.00621 s−1 = A.e-18.58

0.00621 s−1 = A.8.53 * 10-9

A = (0.00621 s−1) / (8.53 * 10-9)

A = 7.28 * 105 s−1 ----------B

Now,

Put new rate constant (k2), Ea and B in A,

0.145 s−1 = (7.28 * 105 s−1).e-(112000 J/mol) / [(8.3145 J/Kmol)* (T2​​​​​​​)]

(0.145 s−1)/(7.28 * 105 s−1) = .e-(112000 J/mol) / [(8.3145 J/Kmol)* (T2​​​​​​​)]

1.99175824 * 10-7 = e-(13470.4432016 K / (T2​​​​​​​)]

Multiply by ln on both side,

ln 1.99175824 * 10-7 = ln e-(13470.4432016 K / (T2​​​​​​​)]

-15.4290778646 = -(13470.4432016 K / (T2​​​​​​​)]

T2​​​​​​​ = (-13470.4432016 K) / (-15.4290778646)

T2​​​​​​​ = 873.06 K

We know that,

873.06 K − 273.15 = 599.91 °C

So,

Rate constant (k2) is 0.145 s−1 at 873.06 K or 599.91 °C [ANSWER]