Question

For the gaseous reaction $3H_2+N_2\iff 2N{H}_3$, the equilibrium pressures of $H_2$ and $N_2$ are $0.4$ atm and $0.8$ atm respectively. The total pressure of the equilibrium system is $2.8$ atm. The value of $Kp$ is:

• A. $50$
• B. $5.0$
• C. $5.0\times {10}^2$
• D. $5.0\times {10}^{-2}$

Answer 1

The correct option is A $50$

$\begin{array}{c}\text{Given equilibrium reaction,}\\ 3H_2+N_2\rightleftharpoons 2{NH}_3\\ \text{t=eqil.}0.4am0.8alm\end{array}$

Total pressure at equilibrium $=2.8$ atm.

Thus, pressure of ${NH}_3=[2\cdot 8-(0.4+0.8\left)\right]$ atm $=1.6$ atm.

Wow, value of kip for an equilibrium $=\frac{{\left[P_C\right]}^{\nu }}{{\left[P_A\right]}^{\alpha }\cdot {\left[P_B\right]}^{\beta }}$, for equilibrium, $\alpha A\left(x\right)+\beta B\left(g\right)\rightleftharpoons \sqrt{C}\left(\vartheta \right)$ Hex, $kp=\frac{(1.6{)}^2}{(0.4{)}^3\cdot (0.8{)}^1}(a+m{)}^{-2}=50(a+m{)}^{-2}$

so, option (A) is correct.