Question

Reaction between nitrogen and oxygen takes place as following: \[2{{N}_{2}}(g)+{{O}_{2}}\rightleftharpoons 2{{N}_{2}}O(g)\]
If a mixture of 0.482 mole of${{N}_{2}}$ and 0.933 mole of ${{O}_{2}}$ is placed in a reaction vessel of volume 10 litre and allowed to form at a temperature for which \[{{\text{K}}_{\text{C}}}\text{ = 2}\text{.0 x 1}{{\text{0}}^{\text{-37}}}\text{ litre mo}{{\text{l}}^{\text{-1}}}\]. Determine the composition of equilibrium mixture.

Answer 1

Hint: To solve the following question, use the formula for calculating the equilibrium constant for the reaction and equate it with the value given in the question. Assign a random variable to the used-up moles at equilibrium.

Complete step by step answer:
Let us look at the reaction given in the question –
\[2{{N}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2{{N}_{2}}O(g)\]
The value for equilibrium constant (${{K}_{c}}$) of this reaction will be –
\[{{K}_{c}}=\dfrac{{{[{{N}_{2}}O]}^{2}}}{{{[{{N}_{2}}]}^{2}}[{{O}_{2}}]}\]
Given in the question,
Equilibrium constant = ${{K}_{c}}$ = \[\text{2}\text{.0 x 1}{{\text{0}}^{\text{-37}}}\text{ litre mo}{{\text{l}}^{\text{-1}}}\]
Volume = 10 L
Initial moles of ${{N}_{2}}$ (dinitrogen gas) = 0.482 mol
Initial moles of ${{O}_{2}}$ (dioxygen gas) = 0.993 mol
- We can calculate concentration by calculating the ratio of mass and volume i.e.
\[concentration=\dfrac{mass}{volume}\]
Concentration of ${{N}_{2}}$ (dinitrogen gas) = \[\dfrac{0.482}{10}mol{{L}^{-1}}\]
Concentration of ${{O}_{2}}$ (dioxygen) = \[\dfrac{0.993}{10}mol{{L}^{-1}}\]
- Now, we need to find the equilibrium concentration of all compounds – nitrogen gas, oxygen gas and nitrous oxide.
Let us assume that ‘x’ oxygen is reacted with ‘2x’ moles of nitrogen. Therefore –
\[\text{2}{{\text{N}}_{\text{2}}}\text{(g) + }{{\text{O}}_{\text{2}}}\rightleftharpoons \text{ 2}{{\text{N}}_{\text{2}}}\text{O(g)}\]

0.0482	0.933	0	Initial moles
\[\text{0}\text{.0482-2x}\]	\[\text{0}\text{.933-x}\]	2x	Moles at equilibrium
\[\dfrac{\text{0}\text{.0482-2x }}{\text{10}}\]	\[\dfrac{\text{0}\text{.933-x}}{\text{10}}\]	\[\dfrac{\text{2x }}{\text{10}}\text{ }\]	Concentration at equilibrium

The value of ${{K}_{c}}$ by substituting above concentration –
\[Kc=\dfrac{{{\left[ \dfrac{\text{2x }}{\text{10}} \right]}^{2}}}{{{\left[ \dfrac{\text{0}\text{.0482-2x }}{\text{10}} \right]}^{2}}\left[ \dfrac{\text{0}\text{.933-x}}{\text{10}} \right]}\]
As we can see, the value of Kc is very small, therefore, the concentration of nitrogen and oxygen that reacted will be very small. So, we will ignore ‘x’ from molar concentration of nitrogen and oxygen.
Therefore, we get –
\[\text{2}\text{.0 x 1}{{\text{0}}^{\text{-37}}}=\dfrac{{{\left[ \dfrac{\text{2x }}{\text{10}} \right]}^{2}}}{{{\left[ \text{0}\text{.0482} \right]}^{2}}\left[ 0.0933 \right]}\]
Solving the above equation –
\[\dfrac{\text{4}{{\text{x}}^{\text{2}}}}{100}={{\left[ \text{0}\text{.0482} \right]}^{2}}\left[ 0.0933 \right]\left[ \text{2}\text{.0 x 1}{{\text{0}}^{\text{-37}}} \right]\]
\[4{{x}^{2}} = 4.34\text{x}{{10}^{-39}}\]
\[{{x}^{2}} = 1.08\text{x}{{10}^{-39}}\]
\[x=\text{3}\text{.3x1}{{\text{0}}^{\text{-20}}}\]
Now, substituting the value of ‘x’ in equilibrium concentrations, we get –
Equilibrium concentration of ${{N}_{2}}$ = \[\dfrac{\text{0}\text{.0482-2x }}{\text{10}} = 0.00482mol{{L}^{-1}}\]
Equilibrium concentration of ${{O}_{2}}$ = \[\dfrac{\text{0}\text{.933-x}}{\text{10}} = 0.0099mol{{L}^{-1}}\]
Equilibrium concentration of ${{N}_{2}}O$ = \[\dfrac{\text{2}x\text{ }}{\text{10}} = 6.6\text{x}{{10}^{-21}}mol{{L}^{-1}}\]
Note: We calculate the equilibrium constant by using the law of mass action. It says that at a constant temperature, the rate at which a substance reacts is always proportional to its active mass.