In a 1 , L container,the reaction of 2 , mol | vedclass

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(B) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.Initial moles: $[N_2] = 2 \, M$,$[H_2] = 5 \, M$,$[NH_3] = 0 \,

Vedclass · Accepted Answer

$7.29 	imes 10^{-3}$

Vedclass · Answer

(B) The balanced chemical equation is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$.Initial moles: $[N_2] = 2 \, M$,$[H_2] = 5 \, M$,$[NH_3] = 0 \, M$.Let the change in concentration of $N_2$ be $x$. At equilibrium:$[N_2] = 2 - x$$[H_2] = 5 - 3x$$[NH_3] = 2x$Given that $[NH_3] = \frac{1}{2} [N_2]$,we have $2x = \frac{1}{2}(2 - x)$.$4x = 2 - x \implies 5x = 2 \implies x = 0.4$.Equilibrium concentrations:$[N_2] = 2 - 0.4 = 1.6 \, M$$[H_2] = 5 - 3(0.4) = 5 - 1.2 = 3.8 \, M$$[NH_3] = 2(0.4) = 0.8 \, M$$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.8)^2}{(1.6)(3.8)^3} = \frac{0.64}{1.6 	imes 54.872} = \frac{0.4}{54.872} \approx 7.29 	imes 10^{-3}$.

In a $1 \, L$ container,the reaction of $2 \, mol$ $N_2$ and $5 \, mol$ $H_2$ occurs. If the equilibrium concentration of $NH_3$ is half the equilibrium concentration of $N_2$,what is the equilibrium constant $(K_c)$?

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For the reaction $N_2 + O_2 \rightleftharpoons 2NO$ at $300 \, ^\circ C$,the value of $K_c$ is $9 \times 10^{-4}$. If equivalent amounts of $N_2$ and $O_2$ are used,what is the concentration of $NO$ at equilibrium (in terms of $a$) (in $, a$)?

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