  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  B
9.32  C
0.66  D
2.68  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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