Question

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?

Answer 1

The equilibrium concentrations are

`[C]=3.3882*10^-3 M`

`[D]=4.3384*10^-4 M`

`[E]=4.3384*10^-4 M`

Explanation:

At equilibrium,

Equation 1
`K_c=([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,

[`C_0`] = [`D_0`] = [`E_0`] = `1.9110*10^-3`

Let's define `C_0=1.9110*10^-3`

At equilibrium

[`C`]=`C_0-x`

[`D`]=`C_0+x`

[`E`]=`C_0+x`

Substituting these three equations into Equation 1 gives

`K_c=((C_0+x)(C_0+x))/(C_0-x)`

Multiply both sides by `C_0-x`

`K_c(C_0-x)=(C_0+x)(C_0+x)=C_0^2+2C_0x+x^2`

Putting this in standard form gives

`x^2+(2C_0+K_c)x+C_0^2-K_cC_0=0`

Solve for `x` using the quadratic formula.

`x=(-(2C_0+K_c)+-sqrt((2C_0+K_c)^2-4*(C_0^2-K_cC_0)))/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 `C_0`. Let's calculate the first value for x:

`x=(-(2*1.9110*10^-3+5.555*10^-5)-sqrt((2*1.9110*10^-3+5.555*10^-5)^2-4*((1.9110*10^-3)^2-5.555*10^-5*1.9110*10^-3)))/2`

This simplifies to `x=-2.400*10^-3 M` which is more moles of `D` or `E` than we started with so this is not physically meaningful.

Lets try the other value

`x=(-(2*1.9110*10^-3+5.555*10^-5)+sqrt((2*1.9110*10^-3+5.555*10^-5)^2-4*((1.9110*10^-3)^2-5.555*10^-5*1.9110*10^-3)))/2`

This simplifies to `x=-1.4772*10^-3 M`

So the equilibrium concentrations are

`[C]=1.9110*10^-3-(-1.4772*10^-3)=3.3882*10^-3 M`

`[D]=1.9110*10^-3+(-1.4772*10^-3)=4.3384*10^-4 M`

`[E]=1.9110*10^-3+(-1.4772*10^-3)=4.3384*10^-4 M`

Note we can check our answer

`K_c=([D][E])/([C])=(4.3384*10^-4)^2/(3.3882*10^-3)=5.555*10^-5`

which is the equilibrium constant given in the problem statement.