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Question

Density of equilibrium mixture of $$N_2O$$ and $$NO_2$$ at 1 atm and 384 K is $$1.84\ g/dm^3$$. Equilibrium constant of the following reaction is: 
$$N_2O_4\rightleftharpoons 2NO_2$$


A
1.98 atm
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B
2.09 atm
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C
2.36 atm
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D
1.48 atm
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Solution

The correct option is B 2.09 atm
The molecular weight
$$\displaystyle M  =\dfrac {\rho RT}{P}= \dfrac {1.84 \times 0.08206 \times 384}{1}=58$$ g/mol.
Now,
$$\displaystyle N_2O_4\rightleftharpoons 2NO_2$$
Let, $$x$$ be the mole fraction of $$\displaystyle N_2O_4 $$. The mole fraction of $$\displaystyle NO_2$$ will be $$\displaystyle 1-x$$
The molecular weights of $$\displaystyle NO_2$$ and $$\displaystyle N_2O_4$$ are 46 g/mol and 92 g/mol respectively.
The average molecular weight of the equilibrium mixture is 58.
$$\displaystyle 58 = 92x+ 46(1-x) $$
$$\displaystyle 58=92x+46-46x$$
$$\displaystyle 58=46+46x$$
$$\displaystyle 12=46x$$
$$\displaystyle x=0.26$$
$$\displaystyle 1-x=1-0.26=0.74$$
The partial pressure is the product of the mole fraction and total pressure (11 atm).
The equilibrium constant $$\displaystyle K_p = \dfrac {P_{NO_2}^2}{P_{N_2O_4}} = \dfrac {(1 \times 0.74)^2}{1 \times 0.26}=2.09$$ atm

Chemistry

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