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# Calculating the Equilibrium Constant: Change in Free Energy

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Survey of physical chemistry

State all of the assumptions (e.g. ideal gas etc).

(previous reaction is 2H??2(g) + O2(g) ----> 2H2O(g))

Given the reaction in the previous question at 298 Kelvin where H2 has a partial pressure of 0.10 bar, O2 is at 0.20 bar, and H2o is at 1.5 bar:
(a) Determine the direction and quantify the change in free energy the reaction must undergo to reach equilibrium.

(b) Calculate the equilibrium constant.

(c) Illustrate the reaction on a plot of Free Energy versus % Reactants/Products (show ?GO, the approximate position of Q for the given partial pressures, and the position of Q at equilibrium).

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Given the reaction:
2H_2(g) + O2(g) ----> 2H_2 O(g)

at 298 kelvin where H_2 has a partial pressure of 0.10 bar, O_2 is at 0.20 bar, and H_2 O is at 1.5 bar:
(a) Determine the direction and quantify the change in free energy the reaction must undergo to reach equilibrium.
(b) Calculate the equilibrium constant.
(c) Illustrate the reaction on a plot of Free Energy versus % Reactants/Products (show GO, the approximate position of Q for the given partial pressures, and the position of Q at equilibrium).

a.) For a reaction with reactants and products not at standard pressures and concentrations, the free energy change is: (please see the attached file)

where (symmetric difference)rGO is the standard free energy change (from thermodynamic tables), R is the gas constant in appropriate units (to match (symmetric difference)rGO's units), T is the temperature in kelvin, and Q is the reaction quotient. (symmetric difference)rGO was found in the previous problem to be approximately -460 kJ mol-1. For the reaction
bB(g) + cC(g) (symmetric difference) dD(g),
Q has the form (please see the attached file)

where fX is the fugacity of species X, and it is divided by the standard reference pressure of one bar to make Q unitless. Here we will assume that all gases involved are ideal, and so their fugacity will be equal to their partial pressures. So at 298 K and at the given partial pressures,

Plugging these (approximate) values into the first equation,

(symmetric ...

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