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Question: Reaction between \[{N_2}\] and \[{O_2}\] takes place as follows: \[2{N_2}\left( g \right) + {O_2}...

Reaction between N2{N_2} and O2{O_2} takes place as follows:
2N2(g)+O2(g)  ⇌2N2O(g)2{N_2}\left( g \right) + {O_2}\left( g \right)\; \rightleftharpoons 2{N_2}O\left( g \right)
If a mixture of 0.482mol0.482mol N2{N_2} and 0.933mol0.933mol of O2{O_2} is placed in a 10L10L reaction vessel and allowed to form N2O{N_2}O at a temperature for which Kc=2.0×10−37Kc = 2.0 \times {10^{ - 37}} , determine the composition of equilibrium mixture.

Explanation

Solution

KC{K_C} is an equilibrium constant which is constant at constant concentrations of the reactants as well as products. It only depends upon the temperature at which the reaction is taking place and the equation of the chemical reaction. KC{K_C} is always written with respect to the concentrations of reactants and products in the chemical reaction.

Complete answer:
We know that KC{K_C} is an equilibrium constant which is constant at constant concentrations of the reactants as well as products. It is always written with respect to the concentrations of reactants and products in the chemical reaction.
The reaction given is:
2N2(g)+O2(g)  ⇌2N2O(g)2{N_2}\left( g \right) + {O_2}\left( g \right)\; \rightleftharpoons 2{N_2}O\left( g \right)
Let the initial concentration of the reactants are:
N2=0.48210{N_2} = \dfrac{{0.482}}{{10}}
O2=0.93310{O_2} = \dfrac{{0.933}}{{10}}
Let xx moles of N2{N_2} get consumed in the reaction. According to the given reaction, x2\dfrac{x}{2} moles of O2{O_2} will react to form xx moles of N2{N_2} . The molar concentration per litre of species before at the equilibrium point is:
N2=0.482−x10{N_2} = \dfrac{{0.482 - x}}{{10}}
O2=0.933−(x2)10{O_2} = \dfrac{{0.933 - \left( {\dfrac{x}{2}} \right)}}{{10}}
N2O=x10{N_2}O = \dfrac{x}{{10}}
Now, the equilibrium constant for the given reaction is written as:
KC=[N2O]2[N2]2[O2]2{K_C} = \dfrac{{{{[{N_2}O]}^2}}}{{{{[{N_2}]}^2}{{[{O_2}]}^2}}}
Kc=2.0×10−37Kc = 2.0 \times {10^{ - 37}} , the value of KC{K_C} is too small and hence, we can neglect xx from the reactant terms because a very small amount of reactants will be consumed in the reaction.
Hence, the value of KC{K_C} becomes:
2.0×10−37=[x10]2[0.48210]2[0.93310]22.0 \times {10^{ - 37}} = \dfrac{{{{\left[ {\dfrac{x}{{10}}} \right]}^2}}}{{{{\left[ {\dfrac{{0.482}}{{10}}} \right]}^2}{{\left[ {\dfrac{{0.933}}{{10}}} \right]}^2}}}
Now, solving the above equation to get the value of x:x:
x=6.6×10−20x = 6.6 \times {10^{ - 20}}
Therefore, the molar concentrations of different species at equilibrium are found to be:
N2=0.0482molL−1{N_2} = 0.0482mol{L^{ - 1}}
O2=0.0933molL−1{O_2} = 0.0933mol{L^{ - 1}}
N2O=6.6×10−20molL−1{N_2}O = 6.6 \times {10^{ - 20}}mol{L^{ - 1}}

Note:
We should remember that there is one more type of equilibrium constant which is written as KP{K_P} . It is constant at constant partial pressures of the reactants as well as products. It only depends upon the temperature at which the reaction is taking place and the equation of the chemical reaction.