Determine the rate constant for each of the following second-order reactions. In each case, express your results in terms of the rate law for the loss of A.

(a) 2 A B + 2 C, given that the concentration of A decreases from 2.74 mmol·L-1 to 1.46 mmol·L-1 in 122 s.

(b) A C + 2 D, given that A0 = 0.320 mol·L-1 and that the concentration of C increases to 0.016 mol·L-1 in 210. s.

You could clarify this by using an arrow. I can't tell where reactants stop and products begin.

a) 2A --> B + 2C

b) A --> C + 2D

To determine the rate constant for each of these second-order reactions, we need to use the integrated rate laws and known concentrations of reactants at different times. First, we will determine the rate law for the loss of A in each case.

(a) For the reaction 2 A -> B + 2 C, the rate law for the loss of A can be expressed as:

Rate = -d[A]/dt = k[A]^2

To find the rate constant (k), we need to rearrange the integrated rate law equation as follows:

1/[A]t - 1/[A]0 = kt

where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, and k is the rate constant.

Given:
[A]0 = 2.74 mmol·L^-1
[A]t = 1.46 mmol·L^-1
t = 122 s

Plugging the known values into the equation, we get:

1/1.46 - 1/2.74 = k(122)

To find k, solve the equation for k:

k = (1/1.46 - 1/2.74) / 122

Calculate the value of k using this equation.

(b) For the reaction A -> C + 2 D, the rate law for the loss of A can be expressed as:

Rate = -d[A]/dt = k[A]

Using the same integrated rate law equation as before:

1/[A]t - 1/[A]0 = kt

Given:
[A]0 = 0.320 mol·L^-1
[A]t = 0.016 mol·L^-1
t = 210 s

Plugging the known values into the equation:

1/0.016 - 1/0.320 = k(210)

To find k, solve the equation for k:

k = (1/0.016 - 1/0.320) / 210

Calculate the value of k using this equation.

By following these steps and performing the calculations, you can determine the rate constant for each of the second-order reactions.