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(D) For the reaction $A_{(g)} \rightleftharpoons B_{(g)}$, the equilibrium constant $K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75$. Initial moles

Vedclass · Accepted Answer

$0.557, 0.418$

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(D) For the reaction $A_{(g)} ightleftharpoons B_{(g)}$, the equilibrium constant $K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75$. Initial moles of $A$ in $1 	ext{ L}$ flask $= 0.5 	ext{ mol}$. Initial moles of $B$ in $1 	ext{ L}$ flask $= 0.375 	ext{ mol}$. After adding $0.1 	ext{ mol}$ of $A$, total moles of $A = 0.5 + 0.1 = 0.6 	ext{ mol}$. Let $x$ be the amount of $A$ reacted to reach new equilibrium. New $[A] = 0.6 - x$ and New $[B] = 0.375 + x$. $K_c = \frac{0.375 + x}{0.6 - x} = 0.75$. $0.375 + x = 0.45 - 0.75x$. $1.75x = 0.075 \Rightarrow x = 0.0428$. New $[A] = 0.6 - 0.0428 = 0.557 	ext{ M}$ and New $[B] = 0.375 + 0.0428 = 0.418 	ext{ M}$.

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