Suppose 0.500 atm H2S, O2, H2O, and SO2 are added to a 1.000 L container at 25 d

Question

Suppose 0.500 atm H2S, O2, H2O, and SO2 are added to a 1.000 L container at 25 degrees Celcius. Calculate the Delta G for this reaction under these conditions.

2H2S(g) + 3O2(g) <=> 2H2O(g) + 2SO2(g) Ka= 3.0X105

Part 2

Calculate the delta S initial for the following raction at 410 K.

CO(g) + H20(g) => CO2(g) + H2(g) Delta H intial = -4.1 x 10^4 Delta S initial universe= 56 J/K

Part 3

The delta G inital for the dissociation of an aqueous solution of a weak acid at 25 degrees celcius is 14.8 kJ. Whaat is the calue of Ka for the acid?

Explanation / Answer

Part 1: %u0394G= %u0394Go + RTlnQ

First we need to calculate %u0394G at standard conditions, that would be %u0394Go. %u0394Go=-RTlnKa = 8.314J/molK*298K*ln(3.0*10^5) = -31245.99282 J/mol.

Next we need to account for the fact that we are not at equilibrium. Our Q value is ([SO2]^2*[H2O]^2)/([O2]^3*[H2S]^2) = 2

RTlnQ = 8.314J/molK * 298 K *ln(2) = 1717.322046 J/mol

%u0394G = -31245.99282 J/mol + 1717.322046 J/mol = -29528.67077J/mol = -29.529 kJ/mol

Part 2: %u0394H = q = -4.1*10^4 (note, you don't have units here but I assume the units are J/mol)

%u0394S = q/T

%u0394S = (-4.1*10^4J/mol)/410 K = -100 J/molK

Part 3:

%u0394Go=-RTlnKa

%u0394Go/(-RT) = lnKa

14.8kJ = 14800 J

lnKa = 14800 J/(-8.314J/K)*(298 K) = -5.973590273

ka = e^ -5.973590273 = .0025450874 = 2.545*10^-3

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