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

6CO_{2}+ 6H_{2}0_{(l)} → C6H_{12}0_{6(s)} ^{+ }60_{2(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^{−1}mol^{−1}

T = 298 K

To find: ∆S_{surr}

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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