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reaction between n2 and o2 ndash takes place as f...

Question:
Reaction between N₂ and O_2– takes place as follows:

2N₂ (g) + O₂ (g) ↔ 2N₂O (g)

If a mixture of 0.482 mol N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_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)

Intial Concentration: 0.482 mol 0.933 mol 0 mol

At equilibrium (0.482-x) mol (0.933 -x) mol x mol

Therefore, at equilibrium, in the 10 L vessel:

N₂ = 0.482-x / 10

O₂ = 0.933-x/2 / 10

N₂O = 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,

N2 = 0.482/10 = 0.0482 molL-1 and O2 = 0.933/10 = 0.0933 molL-1

Now,

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