At T(K),K_c for the reaction A_{2(g)} rightle | vedclass

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(D) Given $K_c = 99.0$. Initial moles of $A_2 = 2 \ mol$,Volume = $1 \ L$,so $[A_2]_{initial} = 2 \ mol \ L^{-1}$. Let $x$ be the concentration of $B_

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$0.02, 1.98$

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(D) Given $K_c = 99.0$. Initial moles of $A_2 = 2 \ mol$,Volume = $1 \ L$,so $[A_2]_{initial} = 2 \ mol \ L^{-1}$. Let $x$ be the concentration of $B_2$ formed at equilibrium. Reaction: $A_{2(g)} \rightleftharpoons B_{2(g)}$ Initial: $2 \ 0$ Equilibrium: $(2-x) \ x$ $K_c = \frac{[B_2]}{[A_2]} = \frac{x}{2-x} = 99.0$. $x = 99(2-x) = 198 - 99x$. $100x = 198 \implies x = 1.98 \ mol \ L^{-1}$. Concentration of $B_2$ at equilibrium = $1.98 \ mol \ L^{-1}$. Concentration of $A_2$ at equilibrium = $2 - 1.98 = 0.02 \ mol \ L^{-1}$. Thus,the concentrations of $A_2$ and $B_2$ are $0.02 \ mol \ L^{-1}$ and $1.98 \ mol \ L^{-1}$ respectively.

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?

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