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Question: At a particular temperature, a \(2.00{\text{L}}\)flask at equilibrium contains \(2.80 \times {10^{ -...

At a particular temperature, a 2.00L2.00{\text{L}}flask at equilibrium contains 2.80×10−42.80 \times {10^{ - 4}} moles of N2{{\text{N}}_{\text{2}}}, 2.5×10−52.5 \times {10^{ - 5}}mole of O2{{\text{O}}_{\text{2}}}and 2.00×10−22.00 \times {10^{ - 2}} mole of N2O{{\text{N}}_{\text{2}}}{\text{O}}. Calculate K{\text{K}} at this temperature for the reaction, 2N2(g)+O2(g)→2N2O(g)2{N_2}(g) + {O_2}(g) \to 2{N_2}O(g)
If [N2]=2×10−4M\left[ {{{\text{N}}_{\text{2}}}} \right] = 2 \times {10^{ - 4}}{\text{M}}, [N2O]=0.200M\left[ {{{\text{N}}_2}{\text{O}}} \right] = 0.200{\text{M}} and [O2]=0.00245M\left[ {{{\text{O}}_2}} \right] = 0.00245{\text{M}}, does this represent a system at equilibrium.

Explanation

Solution

To answer this question, you must recall the formula of equilibrium constant of a reaction in terms of the concentration of reactants, that is Kc{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:
Kc=[N2O]2[O2][N2]2{{\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 Kc{{\text{K}}_{\text{c}}}is equilibrium constant of the reaction
[N2O]\left[ {{{\text{N}}_{\text{2}}}{\text{O}}} \right]is the concentration of N2O{{\text{N}}_{\text{2}}}{\text{O}} at equilibrium
[N2]\left[ {{{\text{N}}_{\text{2}}}} \right]is the concentration of N2{{\text{N}}_{\text{2}}} at equilibrium
[O2]\left[ {{{\text{O}}_{\text{2}}}} \right] is the concentration of O2{{\text{O}}_{\text{2}}} at equilibrium

Complete step by step answer:
For the reaction,
2N2(g)+O2(g)→2N2O(g)2{N_2}(g) + {O_2}(g) \to 2{N_2}O(g)
We are given, volume of flask as2.00L2.00{\text{L}}.
We are given the moles of the reactants and products. So we divide by the volume to get the concentration.
[N2]=2.80×10−42=1.40×10−4M\left[ {{{\text{N}}_{\text{2}}}} \right] = \dfrac{{2.80 \times {{10}^{ - 4}}}}{2} = 1.40 \times {10^{ - 4}}{\text{M}}
[O2]=2.50×10−52=1.25×10−5M\left[ {{{\text{O}}_{\text{2}}}} \right] = \dfrac{{2.50 \times {{10}^{ - 5}}}}{2} = 1.25 \times {10^{ - 5}}{\text{M}}
[N2O]=2.00×10−22=1.00×10−2M\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
Kc=[N2O]2[O2][N2]2{{\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}}}}}
Kc=(1.00×10−2)2(1.25×10−5)(1.40×10−4)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}}}}}
∴Kc=4.08×108\therefore {{\text{K}}_{\text{c}}} = 4.08 \times {10^8}
As Kc>>1{{\text{K}}_{\text{c}}} > > 1, the reaction favors product formation.
For the given values of reactants and products, we must write
Kc’=(0.2)2(2×10−4)2(0.00245)=0.04×1080.0098{{\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}}
∴Kc’=4.08×108\therefore {{\text{K}}_{\text{c}}}{\text{'}} = 4.08 \times {10^8}
As we can clearly see, Kc = Kc’{{\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
Keq=(aC)c(aA)a(aB)b{K_{eq}} = \dfrac{{{{\left( {{a_C}} \right)}^c}}}{{{{\left( {{a_A}} \right)}^a}{{\left( {{a_B}} \right)}^b}}} for a reaction: aA+bB⇌cCaA + bB \rightleftharpoons cC
If K>1K > 1 then equilibrium favours the formation of products, reaction proceeds in the forward direction.
If K<1K < 1 then equilibrium favours the formation of reactants, reaction proceeds in the backward direction.