The dissociation of SO_{3(g)} into SO_{2(g)} | vedclass

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(A) The balanced chemical equation for the dissociation of $SO_{3(g)}$ is:$2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$From the graph,at equil

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$\frac{3^{\frac{3}{2}}}{2^{\frac{3}{2}}}$

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(A) The balanced chemical equation for the dissociation of $SO_{3(g)}$ is:$2SO_{3(g)} ightleftharpoons 2SO_{2(g)} + O_{2(g)}$From the graph,at equilibrium:$P_{SO_3} = 2 \ atm$$P_{SO_2} = 3 \ atm$$P_{O_2} = 1.5 \ atm$ (The curve for $O_2$ levels off at $1.5$)The expression for the equilibrium constant $K_P$ is:$K_P = \frac{(P_{SO_2})^2 	imes (P_{O_2})}{(P_{SO_3})^2}$Substituting the equilibrium partial pressures:$K_P = \frac{(3)^2 	imes (1.5)}{(2)^2} = \frac{9 	imes 1.5}{4} = \frac{13.5}{4} = 3.375$Checking the options,$3.375 = \frac{27}{8} = (\frac{3}{2})^3 = \frac{3^3}{2^3} = \frac{3^{1.5}}{2^{1.5}}$.Thus,$x = \frac{3^{1.5}}{2^{1.5}}$.

The dissociation of $SO_{3(g)}$ into $SO_{2(g)}$ and $O_{2(g)}$ is carried out in a closed container at a constant temperature $T$. The equilibrium constant for the reaction is $K_P = x \ atm$. The partial pressure variation for the three gases is as shown in the graph. The value of $x$ is

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