Question

Question: A 1.00-L vessel at 400 °C contains the following equilibrium concentrations: N2, 1.00M; H2, 0.50M; and NH3, 0.25M. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1M?

Short answer

Since initial concentration of hydrogen is \(0.50{\rm{\;mol,\;}}\)and we have to remove \(0.634{\rm{mol}}\) the equilibrium concentration cannot be \(1.1{\rm{M}}\)

Step-by-step solution

Definition

Chemical equilibrium is the state in which concentration of both thereactants andproducts have no further tendency to change with time.

Nitrogen is thechemical element with thesymbol N andatomic number 7. Nitrogen is anonmetal and the lightest member ofgroup 15 of the periodic table, often called the pnictogens.

Hydrogen is thechemical element with thesymbol H andatomic number 1. Hydrogen is the lightest element. Atstandard conditions hydrogen is agas ofdiatomic molecules having theformula H₂.

Calculate the equilibrium constant

Given information

The volume of the vessel is 1.00L

At equilibrium constants

\(\begin{array}{*{20}{c}}{\left( {{{\rm{N}}_2}} \right) = 1.00{\rm{M}}}\\{\left( {{{\rm{H}}_2}} \right) = 0.50{\rm{M}}}\\{\left( {{\rm{N}}{{\rm{H}}_3}} \right) = 0.25{\rm{M}}}\end{array}\)

The Equillibrium constant is

\({K_c} = \frac{{{{\left| {{\rm{N}}{{\rm{H}}_3}} \right|}^2}}}{{\left| {{N_2}} \right| \times {{\left| {{H_2}} \right|}^3}}}\)

\(\frac{{{{(0.25)}^2}}}{{1.00 \times {{(0.5)}^3}}} = 0.5\)

We have to calculate the number of moles of hydrogen to be removed from the vessel to increase the concentration of nitrogen to 1.1M

The New equilibrium constants

The new equilibrium concentration of the\(\left| {{{\rm{N}}_2}} \right|{\rm{\;is\;}}1.1{\rm{M}}\)

And new initial concentrations are

\(\begin{array}{*{20}{c}}{ - \left( {{{\rm{N}}_2}} \right) = 1.00{\rm{M}}}\\{ - \left( {{{\rm{H}}_2}} \right) = 0.50{\rm{M}} - {\rm{x}}}\\{ - \left( {{\rm{N}}{{\rm{H}}_3}} \right) = 0.25{\rm{M}}}\end{array}\)

\({{\rm{N}}_2}({\rm{g}}) + 3{{\rm{H}}_2}({\rm{g}}) \to 2{\rm{N}}{{\rm{H}}_3}({\rm{g}})\)

Initial(M)	\(1.00\)	\(0.50 - x\)	\(0.25\)
Change(M)	\( + 0.1\)	\( + 0.3\)	\( - 0.2\)
Equillibrium(M)	\(1.1\)	\(0.80 - x\)	\(0.05\)

Calculate the value of X

\(\begin{array}{*{20}{c}}{{K_c} = \frac{{{{\left( {N{H_3}} \right)}^2}}}{{\left( {{N_2}} \right) \times {{\left( {{H_2}} \right)}^3}}}}\\{0.5 = \frac{{{{(0.05)}^2}}}{{1.1 \times {{(0.80 - x)}^3}}}}\end{array}\)

\(\begin{array}{*{20}{c}}{{{(0.80 - x)}^3} = 0.0045}\\{0.80 - x = 0.16565}\\{x = 0.634{\rm{M}}}\end{array}\)

Since initial concentration of hydrogen is \(0.50{\rm{\;mol,\;}}\)and we have to remove \(0.634{\rm{mol}}\) the equilibrium concentration cannot be \(1.1{\rm{M}}\)