Question

What is the value of ΔS_surr for the following reaction at 298 K –

6CO_2(g)  + 6H₂O_(I) → C₆H₁₂O_6(s)  + 6O_2(g)

Given that : 

ΔG⁰ = 2879 KJ mol^-1

ΔS⁰ = -210 J K^-1 mol^-1

Answer 1

Given : 

ΔG⁰ = 2879 KJ mol^-1

= 2879 × 10³ J mol^-1

ΔS⁰ = -210 JK^-1 mol^-1

ΔS_surr = ?

ΔG⁰ = ΔH⁰ – TΔS⁰

∴ ΔH⁰ = ΔG⁰ + TΔS⁰

= 2879 × 10³ + 298 × (-210)

= 2879 × 10³ – 62580

= 2816420 J

Since, for a system, ΔH⁰ is +2816420 J, the surrounding loses heat to system,

∴ ΔH⁰_surr = – 2816420 J

∴  ΔS⁰_surr = \(\frac{ΔH^0_{surr}}{T}\)

= \(\frac{-2816420}{298}\)

= -9451 JK^-1

= -9.451 kJ K^-1

∴ ΔS_surr = -9.451 kJ K^-1