Question

. For 2A + 3B → 2C, [A]0 = 0.15, [B]0 = 0.12, and the initial rate is 0.10 mol/Ls. Assume that the reaction described above is first order in both reactants A and B. Calculate the value of the reaction rate constant.

Answer 1

The rate constant k is ;

K = 2572 L^4/mol^-4•s

Step-by-step explanation:

In this question, we are asked to calculate the rate constant given the chemical equation above.

First, we need to write a complete expression for the rate law. The reaction is a second order type.

We write the rate law as follows;

Rate = k * [A]^2 * [B]^3

Where k is the rate constant and the square brackets indicate the concentrations of both A and B. Let’s input the values :;

0.1 = k * (0.15)^2 * (0.12)^3

0.1 = k * 0.0225 * 0.001728

0.1 = k * 0.00003888

K = 0.1/0.00003888

K = 2572 L^4/mol^-4•s

Answer 2

0.25

Step-by-step explanation:

The rate law for the reaction:

2A(g) + 2B(g) ⟶ 2A(g) + 2B(g)

has been determined to be rate = k[2A]2[2B]. What are the orders with respect to each reactant, and what is the overall order of the reaction?

Answer:

order in 2A= 0.15; order in 2B = 0.10; overall order = 0.25