In the soilution below why is the value of K 6.7478 and not 6.7478 * 1015 Question Calculate ΔG° and K (at 298K) for this reaction: 2H2S(g) + SO2(g) ↔ 3S(s) + 2H2O(g) Expert Answer Step 1 The standard Gibbs free energy can be calculated as follows: ∆G∘=nΣ∆G∘(Products)−mΣ∆G∘(Reactants) where ΔGo: Standard Gibbs free energy n and m are stoichiometric coefficients and Equilibrium Constant (K) can be calculated as follows: ∆G∘=−RTlnK where R: Gas Constant (8.314 JK-1mol-1) ΔGo: Standard Gibbs free energy K: Equilibrium Constant T: Temperature (298 K) Step 2 The standard Gibbs free energy of formation for the following substances are: ΔGof (H2S) = -33.4 kJ/mol ΔGof (SO2) = -300.1 kJ/mol ΔGof (S) = 0 kJ/mol ΔGof (H2O) = -228.6 kJ/mol Substituting all these values in this equation- ∆G∘======[3∆Gf∘(S)+2∆Gf∘(H2O)]−[2∆Gf∘(H2S)+1∆Gf∘(SO2)][3(0 kJ/mol)+2(−228.6 kJ/mol)]−[2(−33.4 kJ/mol)+1(−300.1 kJ/mol)][0 −457.2] kJ/mol−[−66.8−300.1] kJ/mol−457.2−[−366.9] kJ/mol−457.2+366.9 kJ/mol−90.3 kJ/mol Now, equilibrium constant (K) can be calculated as: ∆G∘=−RTlnK substitute all values in this equation as- −90.3 kJ/mol−90.3 kJ/mollnKlnKKK======−8.314×10−3kJK−1 mol−1×298 K×lnK−2.4775 kJ/mol×lnK−90.3 kJ/mol−2.4775 kJ/mol36.4480e36.44806.7478 Step 3 The value of ΔGo is -90.3 kJ/mol and the K value is 6.7478 for this reaction.
In the soilution below why is the value of K 6.7478 and not 6.7478 * 1015
Calculate ΔG° and K (at 298K) for this reaction:
2H2S(g) + SO2(g) ↔ 3S(s) + 2H2O(g)
Expert Answer
The standard Gibbs free energy can be calculated as follows:
∆G∘=nΣ∆G∘(Products)−mΣ∆G∘(Reactants)
where
ΔGo: Standard Gibbs free energy
n and m are
and Equilibrium Constant (K) can be calculated as follows:
∆G∘=−RTlnK
where
R: Gas Constant (8.314 JK-1mol-1)
ΔGo: Standard Gibbs free energy
K: Equilibrium Constant
T: Temperature (298 K)
The standard Gibbs free energy of formation for the following substances are:
ΔGof (H2S) = -33.4 kJ/mol
ΔGof (SO2) = -300.1 kJ/mol
ΔGof (S) = 0 kJ/mol
ΔGof (H2O) = -228.6 kJ/mol
Substituting all these values in this equation-
∆G∘======[3∆Gf∘(S)+2∆Gf∘(H2O)]−[2∆Gf∘(H2S)+1∆Gf∘(SO2)][3(0 kJ/mol)+2(−228.6 kJ/mol)]−[2(−33.4 kJ/mol)+1(−300.1 kJ/mol)][0 −457.2] kJ/mol−[−66.8−300.1] kJ/mol−457.2−[−366.9] kJ/mol−457.2+366.9 kJ/mol−90.3 kJ/mol
Now, equilibrium constant (K) can be calculated as:
∆G∘=−RTlnK
substitute all values in this equation as-
−90.3 kJ/mol−90.3 kJ/mollnKlnKKK======−8.314×10−3kJK−1
mol−1×298 K×lnK−2.4775 kJ/mol×lnK−90.3 kJ/mol−2.4775 kJ/mol36.4480e36.44806.7478
The value of ΔGo is -90.3 kJ/mol and the K value is 6.7478 for this reaction.
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