**Given :**

ΔG^{0} = 2879 KJ mol^{-1}

= 2879 × 10^{3} J mol^{-1}

ΔS^{0} = -210 JK^{-1} mol^{-1}

ΔS_{surr} = ?

ΔG^{0} = ΔH^{0} – TΔS^{0}

∴ ΔH^{0} = ΔG^{0} + TΔS^{0}

= 2879 × 10^{3} + 298 × (-210)

= 2879 × 10^{3 }– 62580

= 2816420 J

**Since, for a system,** **ΔH**^{0} is +2816420 J, the surrounding loses heat to system,

∴ ΔH^{0}_{surr} = – 2816420 J

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

= \(\frac{-2816420}{298}\)

= -9451 JK^{-1}

= -9.451 kJ K^{-1}

∴ ΔS_{surr} = -9.451 kJ K^{-1}