Question

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

6CO₂+ 6H₂0_(l) → C6H₁₂0_6(s) ⁺ 60_2(g),

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

Answer 1

Given:

∆G° = 2879 kJ/mol

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