At 1050 K,for the chemical reaction FeO_{(s) | vedclass

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(D) The reaction is $FeO_{(s)} + CO_{(g)} \rightleftharpoons Fe_{(s)} + CO_{2_{(g)}}$.$Q_p = \frac{P_{CO_2}}{P_{CO}} = \frac{0.8}{1.6} = 0.5$.Since $Q

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$1.92 \ atm, 0.48 \ atm$

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(D) The reaction is $FeO_{(s)} + CO_{(g)} \rightleftharpoons Fe_{(s)} + CO_{2_{(g)}}$.$Q_p = \frac{P_{CO_2}}{P_{CO}} = \frac{0.8}{1.6} = 0.5$.Since $Q_p > K_p$ $(0.5 > 0.25)$,the reaction proceeds in the backward direction.For the backward reaction $Fe_{(s)} + CO_{2_{(g)}} \rightleftharpoons FeO_{(s)} + CO_{(g)}$,$K_p' = \frac{1}{K_p} = \frac{1}{0.25} = 4$.Let $x$ be the change in pressure at equilibrium.$t = 0$: $P_{CO_2} = 0.8 \ atm$,$P_{CO} = 1.6 \ atm$.$t = t_{eq}$: $P_{CO_2} = (0.8 - x)$,$P_{CO} = (1.6 + x)$.$K_p' = \frac{P_{CO}}{P_{CO_2}} \Rightarrow 4 = \frac{1.6 + x}{0.8 - x}$.$3.2 - 4x = 1.6 + x$ $\Rightarrow 5x = 1.6$ $\Rightarrow x = 0.32$.Equilibrium $P_{CO} = 1.6 + 0.32 = 1.92 \ atm$.Equilibrium $P_{CO_2} = 0.8 - 0.32 = 0.48 \ atm$.

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