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?

1 Answer
Apr 28, 2016

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

#"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).