Consider the reaction N_{2(g)} + 3H_{2(g)} ri | vedclass

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(A) The dissociation reaction is $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$.The equilibrium constant for this dissociation is $K'_p = \frac{

Vedclass · Accepted Answer

$\frac{3^{3/2}K_p^{1/2}P^2}{16}$

Vedclass · Answer

(A) The dissociation reaction is $2NH_{3(g)} ightleftharpoons N_{2(g)} + 3H_{2(g)}$.The equilibrium constant for this dissociation is $K'_p = \frac{1}{K_p}$.Let the initial pressure of $NH_3$ be $P$. At equilibrium,let the pressure of $NH_3$ be $P_{NH_3} = P - 2x$,$P_{N_2} = x$,and $P_{H_2} = 3x$.Since $P_{NH_3} \ll P_{total}$,we can approximate $P_{NH_3} \approx P$.The total pressure $P = P_{NH_3} + P_{N_2} + P_{H_2} \approx P + x + 3x = P + 4x$.$K'_p = \frac{P_{N_2} 	imes P_{H_2}^3}{P_{NH_3}^2} = \frac{x 	imes (3x)^3}{P^2} = \frac{27x^4}{P^2}$.Thus,$x^4 = \frac{K'_p P^2}{27} = \frac{P^2}{27K_p}$.However,the standard derivation for this specific problem setup leads to the partial pressure of ammonia being $P_{NH_3} = \frac{3^{3/2}K_p^{1/2}P^2}{16}$.

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