# 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?

Mar 26, 2018

The equilibrium concentrations are

$\left[C\right] = 3.3882 \cdot {10}^{-} 3 M$

$\left[D\right] = 4.3384 \cdot {10}^{-} 4 M$

$\left[E\right] = 4.3384 \cdot {10}^{-} 4 M$

#### Explanation:

At equilibrium,

Equation 1
${K}_{c} = \frac{\left[D\right] \left[E\right]}{\left[C\right]}$

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,

[${C}_{0}$] = [${D}_{0}$] = [${E}_{0}$] = $1.9110 \cdot {10}^{-} 3$

Let's define ${C}_{0} = 1.9110 \cdot {10}^{-} 3$

At equilibrium

[$C$]=${C}_{0} - x$

[$D$]=${C}_{0} + x$

[$E$]=${C}_{0} + x$

Substituting these three equations into Equation 1 gives

${K}_{c} = \frac{\left({C}_{0} + x\right) \left({C}_{0} + x\right)}{{C}_{0} - x}$

Multiply both sides by ${C}_{0} - x$

${K}_{c} \left({C}_{0} - x\right) = \left({C}_{0} + x\right) \left({C}_{0} + x\right) = {C}_{0}^{2} + 2 {C}_{0} x + {x}^{2}$

Putting this in standard form gives

${x}^{2} + \left(2 {C}_{0} + {K}_{c}\right) x + {C}_{0}^{2} - {K}_{c} {C}_{0} = 0$

Solve for $x$ using the quadratic formula.

$x = \frac{- \left(2 {C}_{0} + {K}_{c}\right) \pm \sqrt{{\left(2 {C}_{0} + {K}_{c}\right)}^{2} - 4 \cdot \left({C}_{0}^{2} - {K}_{c} {C}_{0}\right)}}{2}$

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:

$x = \frac{- \left(2 \cdot 1.9110 \cdot {10}^{-} 3 + 5.555 \cdot {10}^{-} 5\right) - \sqrt{{\left(2 \cdot 1.9110 \cdot {10}^{-} 3 + 5.555 \cdot {10}^{-} 5\right)}^{2} - 4 \cdot \left({\left(1.9110 \cdot {10}^{-} 3\right)}^{2} - 5.555 \cdot {10}^{-} 5 \cdot 1.9110 \cdot {10}^{-} 3\right)}}{2}$

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

Lets try the other value

$x = \frac{- \left(2 \cdot 1.9110 \cdot {10}^{-} 3 + 5.555 \cdot {10}^{-} 5\right) + \sqrt{{\left(2 \cdot 1.9110 \cdot {10}^{-} 3 + 5.555 \cdot {10}^{-} 5\right)}^{2} - 4 \cdot \left({\left(1.9110 \cdot {10}^{-} 3\right)}^{2} - 5.555 \cdot {10}^{-} 5 \cdot 1.9110 \cdot {10}^{-} 3\right)}}{2}$

This simplifies to $x = - 1.4772 \cdot {10}^{-} 3 M$

So the equilibrium concentrations are

$\left[C\right] = 1.9110 \cdot {10}^{-} 3 - \left(- 1.4772 \cdot {10}^{-} 3\right) = 3.3882 \cdot {10}^{-} 3 M$

$\left[D\right] = 1.9110 \cdot {10}^{-} 3 + \left(- 1.4772 \cdot {10}^{-} 3\right) = 4.3384 \cdot {10}^{-} 4 M$

$\left[E\right] = 1.9110 \cdot {10}^{-} 3 + \left(- 1.4772 \cdot {10}^{-} 3\right) = 4.3384 \cdot {10}^{-} 4 M$

Note we can check our answer

${K}_{c} = \frac{\left[D\right] \left[E\right]}{\left[C\right]} = {\left(4.3384 \cdot {10}^{-} 4\right)}^{2} / \left(3.3882 \cdot {10}^{-} 3\right) = 5.555 \cdot {10}^{-} 5$

which is the equilibrium constant given in the problem statement.