2X_{6(g)} rightleftharpoons 4X_{3(g)}A 20 L | vedclass

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(A) The equilibrium reaction is $2X_{6(g)} \rightleftharpoons 4X_{3(g)}$.The expression for the equilibrium constant $K_p$ is given by $K_p = \frac{(P

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$5 	imes 10^{-8} \ atm$

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(A) The equilibrium reaction is $2X_{6(g)} ightleftharpoons 4X_{3(g)}$.The expression for the equilibrium constant $K_p$ is given by $K_p = \frac{(P_{X_3})^4}{(P_{X_6})^2}$.Given that the total pressure $P_{total} = P_{X_3} + P_{X_6} = 10 \ atm$ and $P_{X_3} \gg P_{X_6}$,we can approximate $P_{X_3} \approx 10 \ atm$.Substituting the values into the $K_p$ expression:$4 	imes 10^{18} = \frac{(10)^4}{(P_{X_6})^2}$Rearranging for $(P_{X_6})^2$:$(P_{X_6})^2 = \frac{10^4}{4 	imes 10^{18}} = 0.25 	imes 10^{-14} = 25 	imes 10^{-16}$Taking the square root:$P_{X_6} = \sqrt{25 	imes 10^{-16}} = 5 	imes 10^{-8} \ atm$.

$2X_{6(g)} \rightleftharpoons 4X_{3(g)}$
$A$ $20 \ L$ box contains $X_6$ and $X_3$ at equilibrium at $1500 \ K$. The equilibrium constant $K_p = 4 \times 10^{18} \ atm$. Assuming $P_{X_3} \gg P_{X_6}$ and the total pressure is $10 \ atm$,the partial pressure of $X_6$ will be:

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$2X_{6(g)} \rightleftharpoons 4X_{3(g)}$$A$ $20 \ L$ box contains $X_6$ and $X_3$ at equilibrium at $1500 \ K$. The equilibrium constant $K_p = 4 \times 10^{18} \ atm$. Assuming $P_{X_3} \gg P_{X_6}$ and the total pressure is $10 \ atm$,the partial pressure of $X_6$ will be: