Question

Gas $A$ (molar mass 16) and gas $B$ (molar mass 4) are confirmed in a vessel at partial pressure of 2 atm each at ${27}^{0}C$. The gases are then allowed to effuse out at ${27}^{0}C$ through a value till the pressure inside the vessel becomes 2 atm after which the value is closed and the gases are heated to ${127}^{0}C$ and allowed to react as $A+B\rightleftharpoons C;K_p=0.1{atm}^{-1}$. The partial pressure of $C$ at equilibrium assuming $C$ is not formed till the effusing process is completed at ${27}^{0}C$ is $X$ atm. $X$ is:

• A. 0.126
• B. 0.171
• C. 0.187
• D. 0.113

Answer 1

The correct option is A 0.126

The rate of effusion is inversely proportional to the molar mass.
Since the molar mass of gas $A$ is higher, it will effuse at a slower rate and more of $A$ will be left.
The ratio of the number of moles of $A$ effused to the number of moles of $B$ effused is $\sqrt{\frac{4}{16}}=0.5$

After effusion, the partial pressures of $A$ and $B$ will be 1.33 atm and 0.67 atm respectively.

Since heated new pressure of A and B is 1.77 atm and 0.89 atm.

Once, the equilibrium is established, the partial pressures of $A,B$ and $C$ will be $1.77-x$, $0.89-xatm$ and $xatm$ respectively.

$K_p=\frac{x}{(1.77-x)(0.89-x)}=0.1$

$10x=1.575-2.66x+x^2$

$x^2-12.66x+1.575=0$

$x=12.534$ or $x=0.126atm$

Hence, $X=0.126$

Hence, the correct option is $\text{A}$