  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  B
2.09 atm  C
2.36 atm  D
1.48 atm  Solution

The correct option is B 2.09 atmThe 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 $$\displaystyle K_p = \dfrac {P_{NO_2}^2}{P_{N_2O_4}} = \dfrac {(1 \times 0.74)^2}{1 \times 0.26}=2.09$$ atmChemistry

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