a first order reaction with an activation energy of Ea=45.7 kj/mol has a rate constant K=0.012 s^-1 at T=40?

Question

T is in degrees . what is the rate constant at t=80 degrees celsius

Answer 1

1) We need the Arrhenius equation:

k1= A exp(-Ea/RT)
where: k1= 0.012 s^-1
Ea=45.7 kJ/mol
T = 40°C = 313.15 K
R = 8.31 kPa L / mol K

We solve for A:

A = k1 / exp(-Ea/RT)

A=0.012 / exp[-45.7 / (8.31)(313.15)]
A = 0.012 / 0.988 =0.0122 s^-1

2) Now, we use again the Arrhenius equation for T = 80°C :

k2 = A exp(-Ea/RT)
where: k2= ?
A =0.0122 s^-1
Ea=45.7 kJ/mol
T = 80°C = 393.15 K
R = 8.31 kPa L / mol K

k = (0.012)exp [(-45.7) / (8.31)(393.15)]

k= 0.0118 s^-1

That's it!

Good luck!