For the ideal gas reaction,X + Y rightlefthar | vedclass

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(B) For the reaction $X + Y \rightleftharpoons Z$ at $P_{T} = 1 \, bar$:Initial moles at equilibrium: $n_{X} = 1, n_{Y} = 3, n_{Z} = 2$. Total moles $

Vedclass · Accepted Answer

$0.633$

Vedclass · Answer

(B) For the reaction $X + Y ightleftharpoons Z$ at $P_{T} = 1 \, bar$:Initial moles at equilibrium: $n_{X} = 1, n_{Y} = 3, n_{Z} = 2$. Total moles $n_{T} = 6$.Mole fractions: $x_{X} = 1/6, x_{Y} = 3/6, x_{Z} = 2/6$.Partial pressures: $P_{X} = 1/6 \, bar, P_{Y} = 3/6 \, bar, P_{Z} = 2/6 \, bar$.$K_{p} = \frac{P_{Z}}{P_{X} 	imes P_{Y}} = \frac{2/6}{(1/6) 	imes (3/6)} = \frac{2/6}{3/36} = \frac{2}{6} 	imes \frac{36}{3} = 4$.When pressure is increased to $P_{T} = 2 \, bar$,the reaction shifts forward to maintain $K_{p} = 4$.Let $x$ be the moles of $X$ reacted. New moles: $n_{X} = 1 - x, n_{Y} = 3 - x, n_{Z} = 2 + x$. Total moles $n_{T} = 6 - x$.$K_{p} = \frac{P_{Z}}{P_{X} 	imes P_{Y}} = \frac{(n_{Z}/n_{T}) 	imes P_{T}}{(n_{X}/n_{T}) 	imes (n_{Y}/n_{T}) 	imes P_{T}^2} = \frac{n_{Z} 	imes n_{T}}{n_{X} 	imes n_{Y} 	imes P_{T}} = 4$.$\frac{(2 + x)(6 - x)}{(1 - x)(3 - x) 	imes 2} = 4 \Rightarrow \frac{(2 + x)(6 - x)}{(1 - x)(3 - x)} = 8$.$12 + 4x - x^2 = 8(3 - 4x + x^2) = 24 - 32x + 8x^2$.$9x^2 - 36x + 12 = 0 \Rightarrow 3x^2 - 12x + 4 = 0$.Using quadratic formula $x = \frac{12 \pm \sqrt{144 - 48}}{6} = \frac{12 \pm \sqrt{96}}{6} = 2 \pm \frac{9.798}{6} = 2 \pm 1.633$.Since $x $n_{X} = 1 - 0.367 = 0.633 \, mol$.

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