Question

At a particular temperature, a $2.00{\text{L}}$flask at equilibrium contains $2.80 \times {10^{ - 4}}$ moles of ${{\text{N}}_{\text{2}}}$, $2.5 \times {10^{ - 5}}$mole of ${{\text{O}}_{\text{2}}}$and $2.00 \times {10^{ - 2}}$ mole of ${{\text{N}}_{\text{2}}}{\text{O}}$. Calculate ${\text{K}}$ at this temperature for the reaction, $2{N_2}(g) + {O_2}(g) \to 2{N_2}O(g)$
If $\left[ {{{\text{N}}_{\text{2}}}} \right] = 2 \times {10^{ - 4}}{\text{M}}$, $\left[ {{{\text{N}}_2}{\text{O}}} \right] = 0.200{\text{M}}$ and $\left[ {{{\text{O}}_2}} \right] = 0.00245{\text{M}}$, does this represent a system at equilibrium.

Answer 1

Hint: To answer this question, you must recall the formula of equilibrium constant of a reaction in terms of the concentration of reactants, that is ${K_c}$. It is given by the concentration at equilibrium of products divided by the concentration at equilibrium of reactants raised to the power of their respective stoichiometric coefficients.
Formula used:
${{\text{K}}_{\text{c}}} = \dfrac{{{{\left[ {{{\text{N}}_{\text{2}}}{\text{O}}} \right]}^{\text{2}}}}}{{\left[ {{{\text{O}}_{\text{2}}}} \right]{{\left[ {{{\text{N}}_{\text{2}}}} \right]}^{\text{2}}}}}$
Where ${{\text{K}}_{\text{c}}}$is equilibrium constant of the reaction
\[\left[ {{{\text{N}}_{\text{2}}}{\text{O}}} \right]\]is the concentration of ${{\text{N}}_{\text{2}}}{\text{O}}$ at equilibrium
$\left[ {{{\text{N}}_{\text{2}}}} \right]$is the concentration of ${{\text{N}}_{\text{2}}}$ at equilibrium
$\left[ {{{\text{O}}_{\text{2}}}} \right]$ is the concentration of ${{\text{O}}_{\text{2}}}$ at equilibrium

Complete step by step answer:
For the reaction,
$2{N_2}(g) + {O_2}(g) \to 2{N_2}O(g)$
We are given, volume of flask as$2.00{\text{L}}$.
We are given the moles of the reactants and products. So we divide by the volume to get the concentration.
$\left[ {{{\text{N}}_{\text{2}}}} \right] = \dfrac{{2.80 \times {{10}^{ - 4}}}}{2} = 1.40 \times {10^{ - 4}}{\text{M}}$
$\left[ {{{\text{O}}_{\text{2}}}} \right] = \dfrac{{2.50 \times {{10}^{ - 5}}}}{2} = 1.25 \times {10^{ - 5}}{\text{M}}$
\[\left[ {{{\text{N}}_{\text{2}}}{\text{O}}} \right] = \dfrac{{2.00 \times {{10}^{ - 2}}}}{2} = 1.00 \times {10^{ - 2}}{\text{M}}\]
At this stage, the equilibrium constant can be written as
${{\text{K}}_{\text{c}}} = \dfrac{{{{\left[ {{{\text{N}}_{\text{2}}}{\text{O}}} \right]}^{\text{2}}}}}{{\left[ {{{\text{O}}_{\text{2}}}} \right]{{\left[ {{{\text{N}}_{\text{2}}}} \right]}^{\text{2}}}}}$
\[{{\text{K}}_{\text{c}}} = \dfrac{{{{\left( {1.00 \times {{10}^{ - 2}}} \right)}^{\text{2}}}}}{{\left( {1.25 \times {{10}^{ - 5}}} \right){{\left( {1.40 \times {{10}^{ - 4}}} \right)}^{\text{2}}}}}\]
\[\therefore {{\text{K}}_{\text{c}}} = 4.08 \times {10^8}\]
As ${{\text{K}}_{\text{c}}} > > 1$, the reaction favors product formation.
For the given values of reactants and products, we must write
${{\text{K}}_{\text{c}}}{\text{'}} = \dfrac{{{{\left( {0.2} \right)}^2}}}{{{{\left( {2 \times {{10}^{ - 4}}} \right)}^2}\left( {0.00245} \right)}} = \dfrac{{0.04 \times {{10}^8}}}{{0.0098}}$
$\therefore {{\text{K}}_{\text{c}}}{\text{'}} = 4.08 \times {10^8}$
As we can clearly see, ${{\text{K}}_{\text{c}}}{\text{ = }}{{\text{K}}_{\text{c}}}{\text{'}}$
Therefore we can say that the system is in equilibrium.

Note:
The numerical value of an equilibrium constant is obtained by allowing a reaction to proceed to equilibrium and then measuring the concentrations of each substance involved in that reaction. Since the concentrations are measured at equilibrium, the equilibrium constant remains the same for a given reaction independent of initial concentrations of the reactants and products. This knowledge allows us to derive a model expression that can serve as a standard for any reaction. This basic standard form of an equilibrium constant is given as
${K_{eq}} = \dfrac{{{{\left( {{a_C}} \right)}^c}}}{{{{\left( {{a_A}} \right)}^a}{{\left( {{a_B}} \right)}^b}}}$ for a reaction: $aA + bB \rightleftharpoons cC$
If $K > 1$ then equilibrium favours the formation of products, reaction proceeds in the forward direction.
If $K < 1$ then equilibrium favours the formation of reactants, reaction proceeds in the backward direction.