A liquid fuel, methanol, can be produced from coal and water. The final reaction

Question

A liquid fuel, methanol, can be produced from coal and water. The final reaction in one process for doingthis conversion is given below. Using the standardenthalpy and entropy changes for this reaction, determinethe equilibrium constant at 25 oC. (Delta)H = - 129kJ/mol, (delta)S = - 333 J/mol. CO(g) + H2(g) >>>> CH3OH(l)
I think the equilibrium constant is defined by Q=((product)^mol) / ((Reactant)^mol) can someone please help me solve this problem step by step ifpossible? A liquid fuel, methanol, can be produced from coal and water. The final reaction in one process for doingthis conversion is given below. Using the standardenthalpy and entropy changes for this reaction, determinethe equilibrium constant at 25 oC. (Delta)H = - 129kJ/mol, (delta)S = - 333 J/mol. CO(g) + H2(g) >>>> CH3OH(l)
I think the equilibrium constant is defined by Q=((product)^mol) / ((Reactant)^mol) can someone please help me solve this problem step by step ifpossible? CO(g) + H2(g) >>>> CH3OH(l)
I think the equilibrium constant is defined by Q=((product)^mol) / ((Reactant)^mol) can someone please help me solve this problem step by step ifpossible?

Explanation / Answer

We Know that G = H - TS                              = -129*1000 J / mol - (25+273)*(-333J/molK )                              = -129000 + 99234                              = -29766 J / mol Also we know that G = -2.303RT log K Where K = Equilibrium constant R = gas constant = 8.314 J / mol - K T = Temperature = 25+273 = 298 K So log K = -G / ( 2.303RT )              = 5.2167            K= 10^5.2167               = 1.647*10^5

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