Are My Free Energy and Equilibrium Constant Calculations Correct?

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SUMMARY

The calculations for the free energy change and equilibrium constant in the formation of methanol and compound AB are confirmed as correct. For methanol, the free energy change (ΔG) was calculated to be -35.92 kJ using the equation ΔG = ΔG* + RTlnQ, where Q was determined to be 0.010786928. For compound AB, the equilibrium constant (K) was found to be 0.1018 at 181°C, derived from the relationship ΔG* = RTlnK.

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Soaring Crane
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Are my final values correct?

1. Determine the free energy change associated with the formation of methanol from carbon monoxide and hydrogen gas if the partial pressure of CO is 1.0700 atm, H2 is 2.2800 atm, and CH3OH is 0.0600 atm at 298K. The standard free energy of formation for methanol is - 24.70 kJ. Give your answer in kilojoules.

CO (g) + 2H2 (g) --> CH3OH (g)

deltaG* = -24.70 kJ = -24,700 J

deltaG = G* + RTlnQ

Q = [(CH3O)]/[(CO)*(H2)^2] = [0.06 atm]/[(2.28 atm)^2*(1.07 atm)]
= 0.010786928

delta G = -24,700 J + (298 K)(8.314)ln(0.010786928)
= -35921.965 J
= -35.92 kJ




2. The compound AB can be formed from A and B according to the following reaction:A2 (g) + B2 (g) --> 2 AB (g) If the free energy of formation for AB is 8.66 kJ/mol, what is the equilibrium constant for the reaction at 181C? Give your answer to four decimal places.


deltaG* = 2 mol(8.66 kJ/mol) = 17.32 kJ = 17,320 J

deltaG* = RTlnK

K = e^-(G/RT)
= e^-(17,320 J/(8.314 J *454.15 K))
= e^-(4.5871)
= 0.10182 = 0.1018 ?




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