In a closed vessel of volume 0.082 , dm^3 at | vedclass

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(D) The decomposition reaction is: $CH_3OH_{(g)} 	o CO_{(g)} + 2H_{2(g)}$Initial moles: $CH_3OH = 0.01 \, mol$,$CO = 0$,$H_2 = 0$At equilibrium (comp

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$18$

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(D) The decomposition reaction is: $CH_3OH_{(g)} 	o CO_{(g)} + 2H_{2(g)}$Initial moles: $CH_3OH = 0.01 \, mol$,$CO = 0$,$H_2 = 0$At equilibrium (complete decomposition): $CH_3OH = 0 \, mol$,$CO = 0.01 \, mol$,$H_2 = 0.02 \, mol$Total moles of gas $n = 0.01 + 0.02 = 0.03 \, mol$Using the ideal gas law $PV = nRT$:$P = \frac{nRT}{V}$Given $R = 0.082 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,$T = 600 \, K$,$V = 0.082 \, L$$P = \frac{0.03 	imes 0.082 	imes 600}{0.082}$$P = 0.03 	imes 600 = 18 \, atm$

In a closed vessel of volume $0.082 \, dm^3$ at $600 \, K$,the complete decomposition of methanol vapor occurs as follows: $CH_3OH_{(g)} \to CO_{(g)} + 2H_{2(g)}$. If $0.01 \, mol$ of methanol is taken initially,what will be the pressure generated in the vessel in $atm$?

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