At 413 K temperature and 100 atm pressure,1 | vedclass

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(A) The chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.Initial moles: $N_2 = 1$,$H_2 = 3$,$NH_3 = 0$.At equilibrium,$2x = 0.50 \

Vedclass · Accepted Answer

$3.594 	imes 10^{-5} \ atm^{-2}$

Vedclass · Answer

(A) The chemical equation is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$.Initial moles: $N_2 = 1$,$H_2 = 3$,$NH_3 = 0$.At equilibrium,$2x = 0.50 \ mol$,so $x = 0.25 \ mol$.Equilibrium moles: $N_2 = 1 - 0.25 = 0.75 \ mol$,$H_2 = 3 - 3(0.25) = 2.25 \ mol$,$NH_3 = 0.50 \ mol$.Total moles at equilibrium = $0.75 + 2.25 + 0.50 = 3.50 \ mol$.Mole fractions: $X_{N_2} = 0.75/3.50 = 3/14$,$X_{H_2} = 2.25/3.50 = 9/14$,$X_{NH_3} = 0.50/3.50 = 1/7$.Partial pressures $(P = 100 \ atm)$: $P_{N_2} = (3/14) 	imes 100$,$P_{H_2} = (9/14) 	imes 100$,$P_{NH_3} = (1/7) 	imes 100$.$K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(100/7)^2}{(300/14)(900/14)^3} = 3.594 	imes 10^{-5} \ atm^{-2}$.

At $413 \ K$ temperature and $100 \ atm$ pressure,$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are heated in a closed vessel. At equilibrium,$0.50 \ mol$ $NH_3$ is present. Find $K_p$.

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