Consider the partial decomposition of A as:2A | vedclass

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(C) The reaction is $2A_{(g)} \rightleftharpoons 2B_{(g)} + C_{(g)}$.Since the reaction starts with pure $A$,the stoichiometry implies that at equilib

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$10/28$

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(C) The reaction is $2A_{(g)} ightleftharpoons 2B_{(g)} + C_{(g)}$.Since the reaction starts with pure $A$,the stoichiometry implies that at equilibrium,the moles (and thus volumes) of $B$ and $C$ will be in a $2:1$ ratio.Given volume of $C = 100 \ mL$,therefore volume of $B = 200 \ mL$.Total volume of mixture = $700 \ mL$.Volume of $A = 700 - (200 + 100) = 400 \ mL$.Since $P \propto n$ (or $V$ at constant $T$ and $P$),the partial pressure of each gas is $P_i = (V_i / V_{total}) 	imes P_{total}$.$P_A = (400/700) 	imes 10 = 40/7 \ atm$.$P_B = (200/700) 	imes 10 = 20/7 \ atm$.$P_C = (100/700) 	imes 10 = 10/7 \ atm$.$K_P = \frac{(P_B)^2 (P_C)}{(P_A)^2} = \frac{(20/7)^2 (10/7)}{(40/7)^2} = \frac{(400/49) 	imes (10/7)}{1600/49} = \frac{4000/343}{1600/49} = \frac{4000}{343} 	imes \frac{49}{1600} = \frac{40}{7 	imes 4} = \frac{10}{28}$.

Consider the partial decomposition of $A$ as:
$2A_{(g)} \rightleftharpoons 2B_{(g)} + C_{(g)}$
At equilibrium,a $700 \ mL$ gaseous mixture contains $100 \ mL$ of gas $C$ at $10 \ atm$ and $300 \ K$. What is the value of $K_P$ for the reaction?

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Consider the partial decomposition of $A$ as:$2A_{(g)} \rightleftharpoons 2B_{(g)} + C_{(g)}$At equilibrium,a $700 \ mL$ gaseous mixture contains $100 \ mL$ of gas $C$ at $10 \ atm$ and $300 \ K$. What is the value of $K_P$ for the reaction?