# Chemistry Help?

The equilibrium constant of a reaction is 4.22e+02 at 393 K and 9.29e+03 at 492 K. Determine the following for this reaction:

Delta Ho= ______kJ/mol

Delta So=______J/mol-K

What is the value of the equilibrium constant at 443 K?

K=

### 1 Answer

- 冷眼旁觀Lv 61 month ago
1. To find ΔH°:

ln(K₁/K₂) = (ΔH°/R) [(1/T₂) - (1/T₁)]

ln(4.22 × 10² / 9.29 × 10³) = (ΔH°/8.314) [ (1/492) - (1/393)]

ΔH° = 8.314 / {[(1/492) - (1/393)] ln(4.22 × 10² / 9.29 × 10³)} J/mol

ΔH° = 50200 J/mol = 50.2 kJ/mol

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2. To find ΔS°:

ΔG° = ΔH° - TΔS° and ΔG° = -RT ln(K)

Hence, ΔH° - T₁ΔS° = -RT₁ ln(K₁) …… {i}

and ΔH° - T₂ΔS° = -RT₂ ln(K₂) …… {ii}

(i) - (ii):

-T₁ΔS° + T₂ΔS = -RT₁ ln(K₁) + RT₂ ln(K₂)

-ΔS°(T₁ - T₂) = -R [T₁ ln(K₁) - T₂ ln(K₂)]

ΔS° = R [T₁ ln(K₁) - T₂ ln(K₂)] / (T₁ - T₂)

ΔS° = 8.314 [393 ln(4.22 × 10²) - 492 ln(9.29 × 10³)] / (393 - 492) J/(mol K)

ΔS° = 178 J/(mol K)

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3. To find K at 443 K:

ΔH° - TΔS° = -RT ln(K)

ln(K) = (TΔS° - ΔH°) / (RT)

At 443 K:

ln(K) = (443 × 178 - 50200) / (8.314 × 443) = 7.78

K = e⁷˙⁷⁸ = 2.39 × 10³

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