The equilibrium constant Kc for the reaction below is is 5.555×10-5 . C<--->D+E The initial composition of the reaction mixture is [C] = [D] = [E] = 1.9110×10-3 M. What is the equilibrium concentration for C, D, and E?

1 Answer
Mar 26, 2018

The equilibrium concentrations are

[C]=3.3882103M

[D]=4.3384104M

[E]=4.3384104M

Explanation:

At equilibrium,

Equation 1
Kc=[D][E][C]

Let x = the extent of reaction at equilibrium. At constant volume, we can make x = the number of moles of C that are consumed in the reaction at equilibrium. Initially,

[C0] = [D0] = [E0] = 1.9110103

Let's define C0=1.9110103

At equilibrium

[C]=C0x

[D]=C0+x

[E]=C0+x

Substituting these three equations into Equation 1 gives

Kc=(C0+x)(C0+x)C0x

Multiply both sides by C0x

Kc(C0x)=(C0+x)(C0+x)=C20+2C0x+x2

Putting this in standard form gives

x2+(2C0+Kc)x+C20KcC0=0

Solve for x using the quadratic formula.

x=(2C0+Kc)±(2C0+Kc)24(C20KcC0)2

Note that this is an equilibrium reaction, so x can be negative as well as positive, but its absolute value cannot be more than C0. Let's calculate the first value for x:

x=(21.9110103+5.555105)(21.9110103+5.555105)24((1.9110103)25.5551051.9110103)2

This simplifies to x=2.400103M which is more moles of D or E than we started with so this is not physically meaningful.

Lets try the other value

x=(21.9110103+5.555105)+(21.9110103+5.555105)24((1.9110103)25.5551051.9110103)2

This simplifies to x=1.4772103M

So the equilibrium concentrations are

[C]=1.9110103(1.4772103)=3.3882103M

[D]=1.9110103+(1.4772103)=4.3384104M

[E]=1.9110103+(1.4772103)=4.3384104M

Note we can check our answer

Kc=[D][E][C]=(4.3384104)23.3882103=5.555105

which is the equilibrium constant given in the problem statement.