28 g of $N_{2}$ and 6 g of $H_{2}$ were kept at $400℃$ in 1 L vessel, the equilibrium mixture contained 27.54 g of $NH_{3}$. The approximate value of $K_{c}$ for the above reaction can be $(in\ mol^{- 2}L^{2})$

Question

Dr. Abubacker · Accepted Answer

(c)
$N_{2}\  + \ \ \ 3H_{2} ightleftharpoons \ \ \ \ 2{NH}_{3}$
1 3 0 initially
$1 - 0.81\ \ \ 3 - 2.43$
$\left( \mathbf{=}0.19 ight)\left( \mathbf{=}0.57 ight)\ \ 1.62$ at equilibrium
Number of moles of $N_{2} = \frac{28}{28} = 1\ mol$
Number of moles of $H_{2} = \frac{6}{2} = 3\ mol$
Number of moles of $NH_{3} = \frac{27.54}{17} = 1.62\ mol$
$\mathbf{	herefore\ }K_{c} = \frac{\left\lbrack NH_{3} ightbrack^{2}}{\left\lbrack N_{2} ightbrack\left\lbrack H_{2} ightbrack^{3}} = \frac{\lbrack 1.62brack^{2}}{\lbrack 0.19brack\lbrack 0.57brack^{3}} = 75$

Q. 28 g of $N_{2}$ and 6 g of $H_{2}$ were kept at $400℃$ in 1 L vessel, the equilibrium mixture contained 27.54 g of $NH_{3}$. The approximate value of $K_{c}$ for the above reaction can be $(in\ mol^{- 2}L^{2})$

Explanation

Reviewed by Dr. Abubacker

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