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?

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.