what will the rate constant be at 276.0 °C?

Question

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy E_a = 59.0 kJ/mol. If the rate constant of this reaction is 6.7 × 10² M^-1•s^-1 at 226.0 °C, what will the rate constant be at 276.0 °C?

Round your answer to 2 significant digits.

Answer 1

Arrhenius equation:

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

k1 = 6.7x10² M^-1s^-1

k2 = ?

Ea = activation energy = 59.0 kJ/mol

R = gas constant = 8.314 J/Kmol = 0.008314 kJ/Kmol (change units to agree with those of Ea)

T1 = 226ºC + 273 = 499K

T2 = 276ºC + 273 = 549K

Solve for k2:

ln (k2/6.7x10² M^-1s^-1) = -(59.0)/0.008314 (1/549 - 1/499)

ln (k2/6.7x10² M^-1s^-1) = -7096 (0.001821 - 0.00200)

ln (k2/6.7x10² M^-1s^-1) = -7096 (-0.000179) = 1.270

(k2/6.7x10² M^-1s^-1) = 3.56

k2 = 2385 M^-1s^-1

Answer 2

The question provides all the information needed to solve for A using the Arrhenius equation. Then you can calculate k for the second part of the question at a different temperature. Make sure to pay attention to units as they all need to match the gas constant, R. You may need to use natural log (ln) to solve the e function. The Arrhenius equation is as follows:

k=A*e^(-Ea/RT) where:

k=rate constant

A=Arrhenius factor which is different for each reaction

e=Euler's number (2.771828)

Ea=activation energy

R=universal gas constant (8.3144 J mol^-1 K^-1)

T=Temperature (in Kelvin to match R)

The question provides all the information needed to solve for A. Then you can calculate k for the second part of the question at a different temperature

6.7 x10^2 M^-1*s^-1=A*e^(-59 kJ or -59000 J/(8.3144*499))

A=670/6.67 x 10^-7=1 x 10^9

Now at 276, solve for k

k=1 x 10^9*e^(-59000/(8.3144*549)

k=2400