28 , g of N_2 and 6 , g of H_2 were kept at 4 | vedclass

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(A) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ Initial moles: $n(N_2) = \frac{28 \, g}{28 \, g/mol} = 1 \, mol$

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$75$

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(A) The balanced chemical equation is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$ Initial moles: $n(N_2) = \frac{28 \, g}{28 \, g/mol} = 1 \, mol$,$n(H_2) = \frac{6 \, g}{2 \, g/mol} = 3 \, mol$,$n(NH_3) = 0 \, mol$ Equilibrium moles of $NH_3 = \frac{27.54 \, g}{17 \, g/mol} = 1.62 \, mol$ According to stoichiometry,$2 \, mol$ of $NH_3$ are formed from $1 \, mol$ of $N_2$ and $3 \, mol$ of $H_2$. So,$1.62 \, mol$ of $NH_3$ are formed from $0.81 \, mol$ of $N_2$ and $2.43 \, mol$ of $H_2$. Equilibrium moles: $n(N_2) = 1 - 0.81 = 0.19 \, mol$,$n(H_2) = 3 - 2.43 = 0.57 \, mol$,$n(NH_3) = 1.62 \, mol$ Since the volume is $1 \, L$,concentration $[X] = n(X) \, mol/L$. $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(1.62)^2}{(0.19)(0.57)^3} \approx \frac{2.6244}{0.19 	imes 0.185} \approx \frac{2.6244}{0.03515} \approx 74.66 \approx 75$.

$28 \, g$ of $N_2$ and $6 \, g$ of $H_2$ were kept at $400 \, ^oC$ in a $1 \, L$ vessel. The equilibrium mixture contained $27.54 \, g$ of $NH_3$. The approximate value of $K_c$ for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is (in $L^2 \, mol^{-2}$):

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