Question
A 250 ml flask and 100 ml flask are separated by a stopcock. At 350 K, the nitric oxide in the larger flask exerts a pressure of 0.4 atm, and the smaller one contains oxygen at 0.8 atm. The gases are mixed by opening the stopcock. The reactions occurring are as follows. 2NO + O2 → 2NO2$\rightleftharpoons$N2O4
The first reaction is complete while the second one is at equilibrium. Assuming all the gases to behave ideally, calculate the KP for the second reaction if the total pressure is 0.3 atm.
Options
Solution
$n_{NO} = \frac{0.4 \times 250}{RT} = \frac{100}{RT}n_{O_{2}} = \frac{0.8 \times 100}{RT} = \frac{80}{RT}$
2NO + O2 → 2NO2 $\rightleftharpoons$ N2O4
Initial moles $\frac{100}{RT}$ $\frac{80}{RT}$ 0 0
Final moles 0 $\frac{30}{RT}$ $\frac{100}{RT} - x$ $\frac{x}{2}$
From question, $\frac{30}{RT} + \frac{100}{RT} - x + \frac{x}{2} = \frac{0.3 \times 350}{RT}$
$\therefore x = \frac{50}{RT}$
Now KP of second reaction
$= \frac{P_{N_{2}O_{4}}}{P_{NO_{2}}^{2}} = \frac{x/2}{\left( \frac{100}{RT} - x \right)} \times \left( \frac{0.3}{\frac{0.3 \times 350}{RT}} \right)^{- 1} = 3.5\text{ atm}$
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