At T K,3 moles of hydrogen and 1 mole of N | vedclass

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(B) The chemical reaction is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.Initially,we have $1 \ mole$ of $N_2$ and $3 \ moles$ of $H_2$.Let $x$ be

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$2$

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(B) The chemical reaction is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$.Initially,we have $1 \ mole$ of $N_2$ and $3 \ moles$ of $H_2$.Let $x$ be the moles of $N_2$ reacted. According to stoichiometry,$3x$ moles of $H_2$ react to form $2x$ moles of $NH_3$.Given that $2x = 1 \ mole$ of $NH_3$ is formed,so $x = 0.5 \ mole$.At equilibrium,the moles of each gas are:$n(N_2) = 1 - 0.5 = 0.5 \ mole$$n(H_2) = 3 - 3(0.5) = 1.5 \ moles$$n(NH_3) = 1 \ mole$Total moles $n_{total} = 0.5 + 1.5 + 1 = 3 \ moles$.Using Dalton's Law,the partial pressure of $N_2$ is $P(N_2) = (n(N_2) / n_{total}) 	imes P_{total}$.$P(N_2) = (0.5 / 3) 	imes 15 \ atm = 0.5 	imes 5 = 2.5 \ atm$.

At $T \ K$,$3 \ moles$ of hydrogen and $1 \ mole$ of $N_2$ are allowed to react to form ammonia. When $1 \ mole$ of ammonia is formed,the total pressure in the vessel is $15 \ atm$. The partial pressure of $N_2$ in the vessel (in $atm$) is (in $.5$)

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