# 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 #"21 L·mol"^"-1""s"^"-1"# . What is the concentration of **C** after 10 min?

The rate constant is **C** after 10 min?

##### 1 Answer

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

#### Explanation:

**Calculating the integrated rate law**

Your chemical equation is

Let's represent the concentrations as

The rate law is then

The stoichiometry is

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

or

We can use the method of partial fractions to get

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

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 #"C"# after 10 s:**

This is already a very long answer, so I leave it as an exercise for the student to calculate the concentration of