The activation energy for a particular reaction is 102 kJ/mol. If the rate constant is 1.35 × 10⁻⁴ s⁻¹ at 308 K, what is the rate constant at 273 K?

Answer 1

We need to use the Arrhenius equation - ln(k_1/k_2 )=-E_a/R(1/T_1 -1/T_2 )

Substituting values into the equation, we get

ln(k_1/k_2 )=-(102 kJ/mol)/(8.314 x 10^(-3) )(1/308-1/273)

ln(k_1/k_2 )=-.1.23 x 10^4*(3.25 x 10^(-3)-3.66 x 10^(-3))

ln(k_1/k_2 )=-1.23 x 10^4*(-.4.16 x 10^(-4) )=5.11

k_1/k_2 =165

((1.35 x 10^(-4) ))/k_2 =165

k_2=(1.35 x 10^(-4))/165=8.11 x 10^(-7) s^(-1)

Since the 2nd temperature was lower than the 1st temperature, the rate constant will be smaller.

Answer 2

Using the Arrhenius equation, you should be able to calculate the rate constant at the new temperature.

ln (k1/k2) = Ea/R (1/T2 - 1/T1)

Since R has units of J/Kmol, you'll want to change Ea to 102,000 J/mol

k1 = 1.35x10-4

k2 = ?

T1 = 308

T2 = 273

R = 8.314 J/mol K

Plug in the values and solve for k2