1 mol N_2 and 3 mol H_2 are heated at 473 | vedclass

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(A) The reaction is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$Initial moles: $N_2 = 1, H_2 = 3, NH_3 = 0$At equilibrium: $N_2 = 1 - x, H_2

Vedclass · Accepted Answer

$7.5 	imes 10^{-6} \ atm^{-2}$

Vedclass · Answer

(A) The reaction is: $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$Initial moles: $N_2 = 1, H_2 = 3, NH_3 = 0$At equilibrium: $N_2 = 1 - x, H_2 = 3 - 3x, NH_3 = 2x$Given $2x = 0.5 \ mol$,so $x = 0.25 \ mol$.Equilibrium moles: $N_2 = 0.75, H_2 = 2.25, NH_3 = 0.5$.Total moles at equilibrium = $0.75 + 2.25 + 0.5 = 3.5 \ mol$.Partial pressures: $P_{N_2} = (0.75/3.5) 	imes 100 = 21.43 \ atm$,$P_{H_2} = (2.25/3.5) 	imes 100 = 64.29 \ atm$,$P_{NH_3} = (0.5/3.5) 	imes 100 = 14.28 \ atm$.$K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(14.28)^2}{(21.43)(64.29)^3} \approx 7.5 	imes 10^{-6} \ atm^{-2}$.

$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are heated at $473 \ K$ and $100 \ atm$ pressure. At equilibrium,the number of moles of $NH_3$ is $0.5 \ mol$. Calculate the equilibrium constant $K_p$ of the given reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

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