Question

Calculate the Delta H^o , Delta S^o Delta S surroundings, Delta S universe and Delta G^o for the following reaction at 25 ^oC:

2 NiS (s) + 3 O2 (g) --- > 2 SO2 (g) + 2 NIO (s)

Delta H^o = - 890.0 kJ

Delta S^o SO2 (g) = 248 J/K mol, Delta S^o NiO (s) = 38 J/K mol, Delta S^o O2 (g) = 205 J/K mol, Delta S^o NiS (s) = 53 J/K mol

Answer 1

Given chemical transformation,

2 NiS (s) + 3 O₂ (g) ---------- > 2 SO₂ (g) + 2 NiO (s)

1) Calculation of ΔG⁰.

ΔG⁰ = ΔH⁰ –TΔS⁰……………………. (1)

Given data : ΔH⁰ = -890 kJ = -890000 J, T = 25 ⁰C = 25 + 273.15 = 298.15 K.

ΔS⁰ = ?, we have to find out.

ΔS⁰ = ∑ S⁰ (products) - ∑ S⁰(reactants)

ΔS⁰ = [S⁰ (products) + S⁰ (products)] - [S⁰(reactants) + S⁰(reactants) ]

ΔS⁰ = [2 x S⁰ (SO₂) + 2 x S⁰ (NiO)] - [2 x S⁰(NiS) + 3 x S⁰(O2) ]

ΔS⁰ = [2 x (248) + 2 x (38)] - [2 x (53)+ 3 x (205]

ΔS⁰ =572 – 721

ΔS⁰ = -149 J.K^-1.mol^-1

With this value and all other known values in eq. (1),

ΔG⁰ = (-890000) – 298.15 x (-149)

ΔG⁰ = (-890000) -44424.35

ΔG⁰ = -934424.35 J

ΔG⁰ = -934.42 kJ

-ve sign for ΔG⁰ indicate that process is spontaneous.

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2) ΔS⁰ Surrounding = ?

For the given reaction, ΔH⁰ = –890 kJ/mol

-ve sign indicate that heat is evolved from the system. Hence heat is absorbed by the surrounding ,

Therefore, Q_rev. = + 890 kJ/mol = +890000 J/mol

Formula,

ΔS⁰ Surrounding = Q_rev / T

ΔS⁰ Surrounding = +890000 / 298.15

ΔS⁰ Surrounding = +2985 J.K^-1.mol^-1

ΔS⁰ Surrounding = + 2.985 kJ.K^-1.mol^-1

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3) ΔS⁰ universe =?

ΔS⁰ universe = ΔS⁰ (surrounding) + ΔS⁰ (system)

Let us put these calculated values,

ΔS⁰ universe = (+2985) + (-149)

ΔS⁰ universe = +2836 J.K^-1.mol^-1.

And fro spontaneous we expect this only that, ΔS⁰ universe = +ve i.e. ΔS⁰ universe > 0.

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