Q88E - Chemistry (OpenStax)

Question

Question: The equilibrium constant (K_c) for this reaction is 1.60 at 990 °C:

 

Calculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H₂, 2.00 mol of CO₂, 0.750 mol of H₂O, and 1.00 mol of CO to a 5.00-L container at 990 °C.

Step-by-Step Solution

Answer

The number of moles of each component in the final equilibrium mixture is

\(\begin{array}{*{20}{c}}\begin{array}{l}{{\rm{n}}_{{{\rm{H}}_2}}} = 0.612{\rm{mol}}\\{{\rm{n}}_{{\rm{C}}{{\rm{O}}_2}}} = 1.612{\rm{mol}}\end{array}\\{{{\rm{n}}_{{{\rm{H}}_2}{\rm{O}}}} = 1.138{\rm{mol}}}\\{{{\rm{n}}_{{\rm{CO}}}} = 1.388{\rm{mol}}}\end{array}\)

Step 1: Determine the reaction quotient:

Given information:

 

The equilibrium constant is\({K_c} = 1.60\)

The number of moles of H₂ is 1.00 mol

The number of moles of CO₂ is 2.00 mol

 The number of moles of H₂O is 0.750 mol

The number of moles of CO is 1.00 mol

The volume of a container is 5.00 L

We have to find the number of moles of each component in the final equilibrium mixture

The concentration of component

\(\begin{array}{*{20}{c}}{\left( {{{\rm{H}}_2}} \right) = \frac{{1.00{\rm{mol}}}}{{5.00{\rm{L}}}} = 0.2{\rm{M}}}\\{\left( {{\rm{C}}{{\rm{O}}_2}} \right) = \frac{{2.00{\rm{mol}}}}{{5.00{\rm{L}}}} = 0.4{\rm{M}}}\\{\left( {{{\rm{H}}_2}{\rm{O}}} \right) = \frac{{0.750{\rm{mol}}}}{{5.00{\rm{L}}}} = 0.15{\rm{M}}}\\{(CO) = \frac{{1.00{\rm{mol}}}}{{5.00{\rm{L}}}} = 0.2{\rm{M}}}\end{array}\)

Now, we will calculate the reaction quotient, to determine in which direction will reaction go

\(\begin{array}{*{20}{c}}{{Q_c} = \frac{{\left( {{{\rm{H}}_2}{\rm{O}}} \right) \times ({\rm{CO}}]}}{{\left( {{{\rm{H}}_2}} \right) \times \left( {{\rm{C}}{{\rm{O}}_2}} \right)}}}\\{ = \frac{{0.15 \times 0.2}}{{0.2 \times 0.4}}}\\{ = 0.375}\end{array}\)

As\({Q_c} < {K_c}\), the equilibrium will shift to the right

Step 2: Determine the change in equilibrium concentration:

We will find the value of x

\(\begin{array}{*{20}{c}}{{K_c} = \frac{{\left( {{{\rm{H}}_2}{\rm{O}}} \right) \times ({\rm{CO}})}}{{\left( {{{\rm{H}}_2}} \right) \times \left( {{\rm{C}}{{\rm{O}}_2}} \right)}}}\\{1.60 = \frac{{(0.15 + x) \times (0.2 + x)}}{{(0.2 - x) \times (0.4 - x)}}}\\{1.60 = \frac{{{x^2} + 0.35x + 0.03}}{{{x^2} - 0.6x + 0.08}}}\\{{x^2} + 0.35x + 0.03 = 1.6{x^2} - 0.96x + 0.128}\\{0 = 0.6{x^2} - 1.31x + 0.098}\\{{\rm{\;Using equation solver, we get\;}}}\\{x = 0.0776{\rm{M}}}\end{array}\)

Step 3: Determine the concentration of each component:

 The equilibrium concentration of components \({{\rm{H}}_2} = 0.2{\rm{M}} - {\rm{x}} = 0.1224{\rm{M}}\)

\(\begin{array}{*{20}{c}}{{\rm{C}}{{\rm{O}}_2} = 0.4{\rm{M}} - {\rm{x}} = 0.3224{\rm{M}}}\\{{{\rm{H}}_2}{\rm{O}} = 0.15{\rm{M}} + x = 0.2276{\rm{M}}}\\{{\rm{CO}} = 0.2{\rm{M}} + {\rm{x}} = 0.2776{\rm{M}}}\end{array}\) 

Therefore, the number of moles of each component in the final equilibrium mixture is

\(\begin{array}{l}{{\rm{n}}_{{{\rm{H}}_2}}} = 0.1224{\rm{M}} \times 5.00{\rm{L}} = 0.612{\rm{mol}}\\{{\rm{n}}_{{\rm{C}}{{\rm{O}}_2}}} = 0.3224{\rm{M}} \times 5.00{\rm{L}} = 1.612{\rm{mol}}\\\begin{array}{*{20}{c}}{{{\rm{n}}_{{{\rm{H}}_2}{\rm{O}}}} = 0.2276{\rm{M}} \times 5.00{\rm{L}} = 1.138{\rm{mol}}}\\{{{\rm{n}}_{{\rm{CO}}}} = 0.2776{\rm{M}} \times 5.00{\rm{L}} = 1.388{\rm{mol}}}\end{array}\end{array}\)