Question

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?

Answer 1

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

`"A + 2B → C + D"`

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

The rate law is then

`(dc)/dt = kab`

The stoichiometry is

`c_0 = 0`

`a = a_0 –c`

`b = b_0 -2c`

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

`(dc)/dt = 2k(c -a_0 )( c - b_0/2)`

or

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

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 `"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`

`(0.025(0.150 -2c))/(0.150(0.025-c)) = e^0.21 = 1.23`

`(0.150-2c)/(0.025-c) = 6×1.23 = 7.40`

`0.150-2c = 7.40(0.025-c) = 0.185-7.40c`

`5.40c = 0.035`

`c = "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 `"C"` after 10 min (600 s).