# The reaction "A + 2B → C +D" is first-order in the concentrations of **A** and **B**. What is the concentration of **C** after 10 s if the initial concentrations of **A** and **B** are 0.025 mol/L and 0.150 mol/L, respectively?

## The rate constant is $\text{21 L·mol"^"-1""s"^"-1}$. What is the concentration of C after 10 min?

Apr 28, 2016

WARNING! LONG ANSWER! We have to derive the integrated rate law for a mixed second order reaction.

#### Explanation:

Calculating the integrated rate law

$\text{A + 2B → C + D}$

Let's represent the concentrations as $a , b$, and $c$.

The rate law is then

$\frac{\mathrm{dc}}{\mathrm{dt}} = k a b$

The stoichiometry is

${c}_{0} = 0$

a = a_0 –c

$b = {b}_{0} - 2 c$

To make the rate law easier to integrate, we rearrange it as follows,

$\frac{\mathrm{dc}}{\mathrm{dt}} = 2 k \left(c - {a}_{0}\right) \left(c - {b}_{0} / 2\right)$

or

$\frac{\mathrm{dc}}{\left(c - {a}_{0}\right) \left(c - {b}_{0} / 2\right)} = 2 k \mathrm{dt}$

We can use the method of partial fractions to get

(dc)/((c – b_0/2)(b_0/2 – a_0)) - (dc)/((c – a_0)(b_0/2 –a_0)) = 2kdt

Although this is messy, it is relatively easy to integrate to give

1/(b_0/2 – a_0)ln((b_0/2 – c)/(b_0/2)) – 1/(b_0/2 – a_0)ln((a_0 –c)/(a_0)) = 2kt

1/(b_0/2-a_0) ln(((b_0/2 – c)a_0)/((a_0 –c)(b_0/2))) = 2kt

ln[((b_0/2 – c)a_0)/((a_0 –c)(b_0/2))] = 2(b_0/2-a_0)kt

This simplifies to the integrated rate law:

color(blue)(|bar(ul(color(white)(a/a) ln[(a_0 (b_0 – 2c))/( b_0 (a_0 –c))] = (b_0-2a_0)kt color(white)(a/a)|)))" "

Concentration of $\text{C}$ after 10 s:

 ln[(a_0 (b_0 – 2c))/( b_0 (a_0 –c))] = (b_0-2a_0)kt

ln[(0.025(0.150 -2c))/(0.150(0.025-c))] = (0.150-0.050)×0.21 ×10 = 0.21

$\frac{0.025 \left(0.150 - 2 c\right)}{0.150 \left(0.025 - c\right)} = {e}^{0.21} = 1.23$

(0.150-2c)/(0.025-c) = 6×1.23 = 7.40

$0.150 - 2 c = 7.40 \left(0.025 - c\right) = 0.185 - 7.40 c$

$5.40 c = 0.035$

$c = \text{0.006 48 mol/L}$

This is already a very long answer, so I leave it as an exercise for the student to calculate the concentration of $\text{C}$ after 10 min (600 s).