Consider the following equilibrium,CO_{(g)} + | vedclass

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(D) The equilibrium reaction is $CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$.At $t=0$,$n_{CO} = 0.1 \ mol$,$n_{H_2} = a \ mol$,$n_{CH_3OH} =

Vedclass · Accepted Answer

$74$

Vedclass · Answer

(D) The equilibrium reaction is $CO_{(g)} + 2H_{2(g)} ightleftharpoons CH_3OH_{(g)}$.At $t=0$,$n_{CO} = 0.1 \ mol$,$n_{H_2} = a \ mol$,$n_{CH_3OH} = 0$.At equilibrium,$n_{CO} = 0.1 - 0.04 = 0.06 \ mol$,$n_{H_2} = a - 2(0.04) = a - 0.08 \ mol$,$n_{CH_3OH} = 0.04 \ mol$.Total moles at equilibrium $n_{total} = 0.06 + a - 0.08 + 0.04 = a + 0.02 \ mol$.Using the ideal gas law $PV = nRT$ at equilibrium:$5 \ bar 	imes 2 \ dm^3 = (a + 0.02) \ mol 	imes 0.08 \ dm^3 \ bar \ K^{-1} \ mol^{-1} 	imes 500 \ K$.$10 = (a + 0.02) 	imes 40$ $\Rightarrow a + 0.02 = 0.25$ $\Rightarrow a = 0.23 \ mol$.Total moles $n_{total} = 0.25 \ mol$.Mole fractions: $X_{CO} = 0.06/0.25$,$X_{H_2} = (0.23-0.08)/0.25 = 0.15/0.25$,$X_{CH_3OH} = 0.04/0.25$.$K_p = \frac{X_{CH_3OH}}{X_{CO} 	imes X_{H_2}^2} 	imes (P_{total})^{-2} = \frac{0.04/0.25}{(0.06/0.25) 	imes (0.15/0.25)^2} 	imes (5)^{-2}$.$K_p = \frac{0.04 	imes 0.25}{0.06 	imes 0.0225 	imes 25} = \frac{0.01}{0.03375} \approx 0.296$ (Wait,re-calculating: $K_p = \frac{0.04 	imes 0.25^2}{0.06 	imes 0.15^2 	imes 25} = \frac{0.04 	imes 0.0625}{0.06 	imes 0.0225 	imes 25} = \frac{0.0025}{0.03375} \approx 0.07407$).$K_p \approx 74 	imes 10^{-3}$.

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