Question

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

Answer 1

Hint: ${K_C}$ 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. ${K_C}$ is always written with respect to the concentrations of reactants and products in the chemical reaction.

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

Note:
We should remember that there is one more type of equilibrium constant which is written as ${K_P}$ . 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.