7.8. Reaction between nitrogen and oxygen takes place as follows: 2N2?(g) + O2?(g) ? 2N2?O(g) If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a reaction vessel of volume 10 L and allowed to form N2O at a temperature for which Kc = 2.0 x 10-37, determine the composition of the equilibrium mixture.

Let x moles of N2 (g) take part in the reaction. According to the equation, x/2 moles of O2 (g) will react to form x moles of N2O (g). The molar concentration per litre of different species before the reaction and at the equilibrium point is: [N2? ] [O2] [N2? O] Initial concentration 0.482? / 10 0.933/ 10 0 Conc. at equilibrium (0.482 – x)? / 10 (0.933 – x/2) / 10 x/ 10 The value of equilibrium constant (2.0 x 10-37) is extremely small. This means that only small amounts of reactants have reacted. Therefore, value of x is extremely small and can be omitted as far as the reactants are concerned. Applying Law of chemical equilibrium, Kc = [N2O]2 / [N2]2 [O2] = Kc = 2.0 x 10-37 = (x/10)2 / [ (0.482/ 10)2 x (0.933 / 10)2] = 0.01 x2 / (2.1676 x 10-4) x2= 43.352 x 10-40 x = 6.6 x 10-20 Therefore, [N2] = 0.0482 mol/L [O2] = 0.0933 mol/L [N2O] = 0.1 x x= 0.1 x 6.6 x 10-20 = = 6.6 x 10-21 mol/L

7.8. Reaction between nitrogen and oxygen takes place as follows:
2N₂?(g) + O₂?(g) ? 2N₂?O(g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a reaction vessel of volume 10 L and allowed to form N₂O at a temperature for which K_c = 2.0 x 10^-37, determine the composition of the equilibrium mixture.

3 Views|Posted 9 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

9 months ago

Let x moles of N₂ (g) take part in the reaction. According to the equation, x/2 moles of O₂ (g) will react to form x moles of N₂O (g). The molar concentration per litre of different species before the reaction and at the equilibrium point is: