The activation energy, Ea, for a particular reaction is 19.4 kJ/mol. If the rate constant at 80 °C is 0.820 M⁻¹s⁻¹, then what is the value of the rate constant at 243 °C? (R = 8.314 J/mol • K)

Answer 1

Use the Arrhenius equation:

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

k1 = 0.820 M^-1s^-1

k2 = ?

T1 = 80C + 273 = 353K

T2 = 243C + 273 = 516

Ea = 19.4 kJ/mol

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

Solve for k2:

ln (k2/0.820) = - 19.4/0.008314 (1/516 - 1/353)

ln (k2/0.820) = - 2333 (0.001938 - 0.002833)

ln (k2/0.820) = 2.088

k2/0.820 = 8.07

k2 = 6.62 M^-1s^-1

Answer 2

To find the value of the rate constant at 243 °C, we can use the Arrhenius equation:

k2 = k1 * e^((Ea/R) * (1/T1 - 1/T2))

where k1 = rate constant at T1 (0.820 M⁻¹s⁻¹) k2 = rate constant at T2 (unknown) Ea = activation energy (19.4 kJ/mol = 19400 J/mol) R = gas constant (8.314 J/mol • K) T1 = initial temperature (80 °C = 353.15 K) T2 = final temperature (243 °C = 516.15 K)

Plugging in the values:

k2 = 0.820 * e^((19400/8.314) * (1/353.15 - 1/516.15))

Now, let's calculate the value of k2:

k2 = 0.820 * e^(2.333 * (0.00283 - 0.00194)) k2 = 0.820 * e^(2.333 * 0.00089) k2 ≈ 0.820 * e^2.076 k2 ≈ 0.820 * 7.977 k2 ≈ 6.541

So the value of the rate constant at 243 °C is approximately 6.541 M⁻¹s⁻¹.