Reaction between \(N_2\) and \(O_2^–\) takes place as follows:
\(2N_2 (g) + O_2 (g) ⇋ 2N_2O (g)\)
If a mixture of \(0.482 \ mol\) \(N_2\) and \(0.933\ mol\) of \(O_2\) is placed in a \(10\ L\) reaction vessel and allowed to form \(N_2O\) at a temperature for which \(K_c = 2.0 × 10^{–37}\), determine the composition of equilibrium mixture.
\(2N_2 (g) + O_2 (g) ⇋ 2N_2O (g)\)
If a mixture of \(0.482 \ mol\) \(N_2\) and \(0.933\ mol\) of \(O_2\) is placed in a \(10\ L\) reaction vessel and allowed to form \(N_2O\) at a temperature for which \(K_c = 2.0 × 10^{–37}\), determine the composition of equilibrium mixture.
Solution and Explanation
Let the concentration of N2O at equilibrium be x.
The given reaction is:
\(2N_2(g) + O_2(g) ↔ 2N_2O(g)\)
Initial conc. \(0.482\ mol\) \(0.933 \ mol\) \(0\)
At equilibrium \((0.482 - x)\ mol\) \((1.933 - x)\ mol\) \(x\) \(mol\)
Therefore, at equilibrium, in the 10 L vessel:
\([N_2] = \frac {0.482 - x }{10},\)
\([O_2] = \frac {0.933 - \frac x2}{10} ,\)
\([N2O] = \frac {x}{10}\)
The value of equilibrium constant i.e., \(K_c= 2.0 × 10^{-37}\) is very small. Therefore, the amount of N2 and O2 reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N2 and O2.
Then, \([N_2] = \frac {0.482}{10} = 0.0482\ mol L^{-1 }\)
and \([O_2] = \frac {0.933}{10}= 0.0933 \ mol L^{-1}\)
Now,
\(K_c = \frac {[N_2O(g)]^2}{[N_2(g)]^2 [O_2(g)] }\)
⇒ \(2.0×10^{-37} = \frac {(\frac {x}{10})^2}{(0.0482)^2 (0.0933) }\)
⇒ \(\frac {x^2}{100} = 2.0×10^{-37}×(0.0482)^2×(0.0933)\)
⇒ \(x^2 = 43.35×10^{- 40}\)
⇒ \(x = 6.6×10^{-20}\)
\([N_2O] = \frac {x}{10}\)
\([N_2O] =\) \(\frac {6.6 × 10^{- 20}}{10}\)
\([N_2O] = 6.6×10^{-21}\)
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Concepts Used:
Equilibrium Constant
The equilibrium constant may be defined as the ratio between the product of the molar concentrations of the products to that of the product of the molar concentrations of the reactants with each concentration term raised to a power equal to the stoichiometric coefficient in the balanced chemical reaction.
The equilibrium constant at a given temperature is the ratio of the rate constant of forwarding and backward reactions.
Equilibrium Constant Formula:
Kequ = kf/kb = [C]c [D]d/[A]a [B]b = Kc
where Kc, indicates the equilibrium constant measured in moles per litre.
For reactions involving gases: The equilibrium constant formula, in terms of partial pressure will be:
Kequ = kf/kb = [[pC]c [pD]d]/[[pA]a [pB]b] = Kp
Where Kp indicates the equilibrium constant formula in terms of partial pressures.
- Larger Kc/Kp values indicate higher product formation and higher percentage conversion.
- Lower Kc/Kp values indicate lower product formation and lower percentage conversion.
Medium Kc/Kp values indicate optimum product formation.
Units of Equilibrium Constant:
The equilibrium constant is the ratio of the concentrations raised to the stoichiometric coefficients. Therefore, the unit of the equilibrium constant = [Mole L-1]△n.
where, ∆n = sum of stoichiometric coefficients of products – a sum of stoichiometric coefficients of reactants.