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(B) The target reaction is the oxidation of $2 	ext{ mol}$ of $NH_{3}$ to $NO$:$2 NH_{3} + \frac{5}{2} O_{2} ightleftharpoons 2 NO + 3 H_{2} O$Give

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$K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$

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(B) The target reaction is the oxidation of $2 	ext{ mol}$ of $NH_{3}$ to $NO$:$2 NH_{3} + \frac{5}{2} O_{2} ightleftharpoons 2 NO + 3 H_{2} O$Given equations:$(i) N_{2} + 3 H_{2} ightleftharpoons 2 NH_{3} ; K_{1}$$(ii) N_{2} + O_{2} ightleftharpoons 2 NO ; K_{2}$$(iii) H_{2} + \frac{1}{2} O_{2} ightleftharpoons H_{2} O ; K_{3}$To obtain the target reaction:$1$. Reverse equation $(i)$: $2 NH_{3} ightleftharpoons N_{2} + 3 H_{2} ; K' = \frac{1}{K_{1}}$$2$. Keep equation $(ii)$ as it is: $N_{2} + O_{2} ightleftharpoons 2 NO ; K_{2}$$3$. Multiply equation $(iii)$ by $3$: $3 H_{2} + \frac{3}{2} O_{2} ightleftharpoons 3 H_{2} O ; K'' = K_{3}^{3}$Adding these three equations:$(2 NH_{3} + N_{2} + O_{2} + 3 H_{2} + \frac{3}{2} O_{2}) ightleftharpoons (N_{2} + 3 H_{2} + 2 NO + 3 H_{2} O)$Simplifying gives: $2 NH_{3} + \frac{5}{2} O_{2} ightleftharpoons 2 NO + 3 H_{2} O$The equilibrium constant $K$ is:$K = K' \cdot K_{2} \cdot K'' = \frac{1}{K_{1}} \cdot K_{2} \cdot K_{3}^{3} = K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$

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At $T(K)$,$K_c$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $99.0$. Two moles of $A_{2(g)}$ were heated to $T(K)$ in a $1 \ L$ closed flask to reach the above equilibrium. What are the concentrations (in $mol \ L^{-1}$) of $A_{2(g)}$ and $B_{2(g)}$ respectively at equilibrium?

In the equilibrium $AB \rightleftharpoons A + B$; if the equilibrium concentration of $A$ is doubled,the equilibrium concentration of $B$ would become:

For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the value of $K_p$ changes with:

For the system $A_{(g)} + 2B_{(g)} \rightleftharpoons C_{(g)}$,the equilibrium concentrations are $[A] = 0.06 \ mol/L$,$[B] = 0.12 \ mol/L$,and $[C] = 0.216 \ mol/L$. The $K_{eq}$ for the reaction is:

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The following equilibrium constants are given:$N_{2} + 3 H_{2} \rightleftharpoons 2 NH_{3} ; K_{1}$$N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$$H_{2} + \frac{1}{2} O_{2} \rightleftharpoons H_{2} O ; K_{3}$The equilibrium constant for the oxidation of $2 \text{ mole}$ of $NH_{3}$ to give $NO$ is