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What is the value of for the following reaction at 298 K - Chemistry

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What is the value of for the following reaction at 298 K -

6CO2+ 6H20(l) → C6H1206(s) + 602(g),

Given that: ΔG° = 2879 kj mol-1, AS = -210 JK-1 mol-1.

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Solution

Given:

∆G° = 2879 kJ/mol

∆S° = −210 J K−1 mol−1
= −0.210 kJ K−1mol−1
T = 298 K

To find: ∆Ssurr

Formula: ∆G° = ∆H° − T∆S°

Calculation: ∆G° = ∆H° − T∆S°

2879 = ∆H° − (298 × −0.210)
2879 = ∆H° − (−62.58)
2879 = ∆H° + 62.58

∆H° = 2816.42 kJ

`DeltaS_(surr)=q_(surr)/T`

`=-(DeltaH^@)/T`

`=-(2816.42)/298`

= − 9.45 kJ/K

 

Concept: Relation Between Gibbs Energy Change and Emf of a Cell
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