Question

The Kp value for the reaction $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ at ${460}^{o}C$ is 49. If the initial pressure of $H_2$ and $I_2$ are 0.5 atm respectively, determine the partial pressure of $H_2,I_2$ and $HI$ gases at equilibrium.

• A. 0.11 atm, 0.11 atm, 0.77 atm
• B. 0.11 atm, 0.77 atm, 0.11 atm
• C. 0.77 atm, 0.11 atm, 0.11 atm
• D. None of these

Answer 1

The correct option is A 0.11 atm, 0.11 atm, 0.77 atm

$H_2+I_2\to 2HI$

Initial : $0.5$ $0.5$ $0$

At eq. : $(0.5-x)$ $(0.5-x)$ $2x$

$K_p=\frac{(2x{)}^2}{(0.5-x{)}^2}=49$

On solving the above equation, $2x=3.5-7x$

$9x=3.5$

$x=0.389$

At equillibrium,Partial pressure of hydrogen, $p^{\prime }{H}_2=0.5-0.389=0.111atm$

Partial pressure of $I_2,p^{\prime }{I}_2=0.5-0.389=0.111atm$

Partial pressure of hydrogen iodide , $p^{\prime }HI=2\times 0.389=0.77atm$

Hence, option A is correct.