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Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

Reaction between N2N_2 and O2–O_2^– takes place as follows:
2N2(g)+O2(g)⇋2N2O(g)2N_2 (g) + O_2 (g) ⇋ 2N_2O (g)
If a mixture of 0.482 mol0.482 \ mol N2N_2 and 0.933 mol0.933\ mol of O2O_2 is placed in a 10 L10\ L reaction vessel and allowed to form N2ON_2O at a temperature for which Kc=2.0×10–37K_c = 2.0 × 10^{–37}, determine the composition of equilibrium mixture.

Answer

Let the concentration of N2O at equilibrium be x.
The given reaction is:
2N2(g)+O2(g)↔2N2O(g)2N_2(g) + O_2(g) ↔ 2N_2O(g)
Initial conc. 0.482 mol0.482\ mol 0.933 mol0.933 \ mol 00
At equilibrium (0.482−x) mol(0.482 - x)\ mol (1.933−x) mol(1.933 - x)\ mol xx molmol
Therefore, at equilibrium, in the 10 L vessel:
[N2]=0.482−x10,[N_2] = \frac {0.482 - x }{10},

[O2]=0.933−x210,[O_2] = \frac {0.933 - \frac x2}{10} ,

[N2O]=x10[N2O] = \frac {x}{10}
The value of equilibrium constant i.e., Kc=2.0×10−37K_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, [N2]=0.48210=0.0482 molL−1[N_2] = \frac {0.482}{10} = 0.0482\ mol L^{-1 }

and [O2]=0.93310=0.0933 molL−1[O_2] = \frac {0.933}{10}= 0.0933 \ mol L^{-1}
Now,
Kc=[N2O(g)]2[N2(g)]2[O2(g)]K_c = \frac {[N_2O(g)]^2}{[N_2(g)]^2 [O_2(g)] }

⇒ 2.0×10−37=(x10)2(0.0482)2(0.0933)2.0×10^{-37} = \frac {(\frac {x}{10})^2}{(0.0482)^2 (0.0933) }

⇒ x2100=2.0×10−37×(0.0482)2×(0.0933)\frac {x^2}{100} = 2.0×10^{-37}×(0.0482)^2×(0.0933)

⇒ x2=43.35×10−40x^2 = 43.35×10^{- 40}

⇒ x=6.6×10−20x = 6.6×10^{-20}

[N2O]=x10[N_2O] = \frac {x}{10}

[N2O]=[N_2O] = 6.6×10−2010\frac {6.6 × 10^{- 20}}{10}

[N2O]=6.6×10−21[N_2O] = 6.6×10^{-21}