What would the rate of the reaction be (in mol L−1 s−1) when concentration of A = 0.43 mol L−1, B = 0.185 mol L−1, and C = 0.207 mol L−1?

Question

An experiment was designed to the study the reaction kinetics of a reaction that contains the reactants A, B, and C. Given the following data:

A₀ (mol L⁻¹) B₀ (mol L⁻¹). C₀ (mol L⁻¹). Initial Rate (mol L⁻¹ s⁻¹)

0.300 0.100 0.150 1.25

0.300 0.100 0.450 11.25

0.900 0.100 0.450 11.25

0.300 0.400 0.450 180.00

What would the rate of the reaction be (in mol L⁻¹ s⁻¹) when concentration of A = 0.43 mol L⁻¹, B = 0.185 mol L⁻¹, and C = 0.207 mol L⁻¹?

Answer 1

First, find the order for each reactant.

Then write the rate equation and solve for k, the rate constant

Finally, use the concentrations given along with the rate equation to solve for the rate.

Line 2 & 3: [A] triples, [B] and [C] remain constant and rate doesn't change. ==> zero order in A

Line 3 and 4: [B] quadruples and [C] remains constant and rate increases 16x => 2nd order in B

Line 1 & 2: [C] triples and [B] remains constant and rate increases 9x => 2nd order in C

Rate law is..

Rate = k[B]²[C]²

Solving for k, rate constant, using line 1, we have...

1.25 M/s = k (0.1 M)²(0.15 M)² = 0.000225 M⁴

k = 1.25/0.000225

k = 5556 M^-3s^-1

When A = 0.43 mol L⁻¹, B = 0.185 mol L⁻¹, and C = 0.207 mol L⁻¹ the rate of reaction is....

rate = 5556(0.185)2(0.207)2 = 8.15 M/s