Answer to Question #105913 in Chemistry for Dianna Zebib

Question

For the reaction A + B  C, the rate constant at 215 oC is 5.0 x 10-3 and the rate constant at 452o C is 1.2 x 10-1.
3

(a) What is the activation energy in kJ/mol? (b) What is the rate constant at 100o C.

Answer 1

1) k₁ = 5.0 * 10^-3;

T₁ = 215 ^oC = 488 K.

2) k₂ = 1.2 * 10^-1 (let us assume that the constant has such a value since, by the condition of the task, it is not completely clear);

T₂ = 452 ^oC = 725 K.

3) k₃ = ???

T₃ = 100 ^oC = 373 K.

Solution (a):

Arrhenius equation looks like this:

𝑘 = 𝐴𝑒^{−𝐸𝑎/𝑅𝑇}

Where:

𝑘 is the rate constant, 𝐴 is the pre–exponential factor, 𝐸_𝑎 is the activation energy, 𝑅 is the ideal gas constant (8.314 𝐽⁄𝐾 ∙ 𝑚𝑜𝑙), and 𝑇 is the absolute temperature of the system.

Then, for two points Arrhenius equation looks like this:

ln(𝑘₂/𝑘₁) = 𝐸_𝑎*(1/𝑇₁ − 1/𝑇₂ ) / R

Then,

ln(1.2 * 10^-1 / 5.0 * 10^-3) = 𝐸_𝑎 * (1/488 − 1/725 ) / 8.314;

ln(24) = 𝐸_𝑎 * 8.057 * 10^-5;

3.178 = 𝐸_𝑎 * 8.057 * 10^-5;

𝐸_𝑎 = 39444.63 𝐽⁄𝑚𝑜𝑙 = 39.44 k𝐽⁄𝑚𝑜𝑙.

Answer (a): 𝐸_𝑎 = 39.44 k𝐽⁄𝑚𝑜𝑙.

Solution (b):

ln(𝑘₂/𝑘₃) = 𝐸_𝑎*(1/𝑇₃ − 1/𝑇₂ ) / R

Then,

ln(1.2 * 10^-1/𝑘₃) = 39444.63 * (1/373 − 1/725 ) / 8.314;

ln(1.2 * 10^-1/𝑘₃) = 6.1755;

ln(1.2 * 10^-1) - ln(𝑘₃) = 6.1755;

-2.12026 - ln(𝑘₃) = 6.1755;

ln(𝑘₃) = -8.29576;

𝑘₃ = e^-8.29576 = 2.49 * 10^-4;

𝑘₃ = 2.49 * 10^-4.

Check:

ln(𝑘₁/𝑘₃) = 𝐸_𝑎*(1/𝑇₃ − 1/𝑇₁ ) / R

Then,

ln(5.0 * 10^-3/𝑘₃) = 39444.63 * (1/373 − 1/488 ) / 8.314;

ln(5.0 * 10^-3/𝑘₃) = 2.9974;

ln(5.0 * 10^-3) - ln(𝑘₃) = 2.9974;

-5.2983 - ln(𝑘₃) = 2.9974;

ln(𝑘₃) = -8.2957;

𝑘₃ = e^-8.2957 = 2.49 * 10^-4;

𝑘₃ = 2.49 * 10^-4. - Correct.

Answer (b): 𝑘₃ = 2.49 * 10^-4.

Answer: (a) 𝐸𝑎 = 39.44 k𝐽⁄𝑚𝑜𝑙; (b): 𝑘₃ (at 100 ^oC) = 2.49 * 10^-4.