The equilibrium concentrations of N_2, H_2 an | vedclass

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(D) The balanced chemical equation for the formation of $NH_3$ is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.First,calculate the equilibrium cons

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$320(RT)^{-2}$

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(D) The balanced chemical equation for the formation of $NH_3$ is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$.First,calculate the equilibrium constant $K_c$:$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(1.6 	imes 10^{-2})^2}{(1.25 	imes 10^{-2}) 	imes (4.0 	imes 10^{-2})^3}$.$K_c = \frac{2.56 	imes 10^{-4}}{(1.25 	imes 10^{-2}) 	imes (64 	imes 10^{-6})} = \frac{2.56 	imes 10^{-4}}{80 	imes 10^{-8}} = \frac{2.56 	imes 10^{-4}}{8 	imes 10^{-7}} = 0.32 	imes 10^3 = 320$.The relationship between $K_p$ and $K_c$ is $K_p = K_c(RT)^{\Delta n}$.Here,$\Delta n = 2 - (1 + 3) = 2 - 4 = -2$.Therefore,$K_p = 320(RT)^{-2}$.

The equilibrium concentrations of $N_2, H_2$ and $NH_3$ in the formation of $NH_3$ at $500 \ K$ are $1.25 \times 10^{-2} \ M, 4.0 \times 10^{-2} \ M$ and $1.6 \times 10^{-2} \ M$ respectively. The equilibrium constant $K_{p}$ at the same temperature is

Similar Questions

The reaction $A(g) \rightleftharpoons B(g) + C(g)$ was initiated with the amount $a$ of $A(g)$. At equilibrium,it is found that the amount of $A(g)$ remaining is $(a-x)$ at a total pressure of $p$. The equilibrium constant $K_{p}$ of the reaction can be calculated from the expression:

For the reaction: $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,$K_P = 2 \times 10^{12}$ at $27^{\circ} C$ and $1 \ atm$ pressure. The $K_C$ for the same reaction is $......... \times 10^{13}$. (Nearest integer)
(Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

For the following gaseous equilibria at $300 \ K$,find the increasing order of the ratio $\frac{K_p}{K_c}$ for $X, Y,$ and $Z$:
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$Y: PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
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