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Question

When 0.1 moles of $$N_2O_4 (g)$$ was placed in a 1.0 litre flask at 400 K, the following equilibrium is reached with a total pressure of 6 bar.

$${N_2O_4} \, \rightleftharpoons \, {2NO_2}$$

Assuming ideal behaviour of the gases, the partial pressure of $$N_2O_4 (g)$$ at equilibrium in bar is:

[R= 0.083 bar L mol$$^{-1} k^{-1}$$]


A
3.32
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B
9.32
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C
0.66
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D
2.68
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Solution

The correct option is B 0.66
$$N_2O_4 \ \ \ \     \longrightarrow \ \ \ \  2NO_2 $$
$$0.1 - x$$                  $$ 2x$$ 

P1 = partial pressure of $$N_2O_4$$
P2 = partial pressure of $$NO_2$$ 

Partial pressure of $$N_2O_4 = 6 \times\cfrac{ (0.1 - x)}  { (0.1 +x )}$$ 

Total moles = $$\cfrac {P V}{ RT}$$ 

Total moles = $$6 \times \cfrac{1}{(0.083 \times 400)}$$  $$= 0.18\ moles$$ 

Total moles $$= 0.1 + x = 0.18 $$

$$x = 0.18 - 0.1 = 0.08$$
 
Partial pressure of $$N_2O_4= 6 \times\cfrac{ ( 0.1 - 0.08)}{ ( 0.1 + 0.08) }$$

Partial pressure of $$N_2O_4 = \cfrac{(0.6 - 0.48)}{ 0.18}$$

Partial pressure of $$N_2O_4 = 0.66\ bar$$

Hence, option C is correct.

Chemistry

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