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