Question

Find the enthalpy change for the

reaction

CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g)

when:

C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol

S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol

C(s) + 2 S(s) → CS2(l); ΔHf = 87.9 kJ/mol​

Answer 1

Explanation:

ENTHALPY OF REACTION

[1ΔHf(CO2 (g)) + 2ΔHf(SO2 (g))] - [1ΔHf(CS2 (ℓ)) + 3ΔHf(O2 (g))]

[1(-393.51) + 2(-296.83)] - [1(89.7) + 3(0)] = -1076.87 kJ

-1,076.87 kJ (exothermic)

ENTROPY CHANGE

[1ΔSf(CO2 (g)) + 2ΔSf(SO2 (g))] - [1ΔSf(CS2 (ℓ)) + 3ΔSf(O2 (g))]

[1(213.68) + 2(248.11)] - [1(151.34) + 3(205.03)] = -56.53 J/K

-56.53 J/K (decrease in entropy)

FREE ENERGY OF REACTION (AT 298.15 K)

From ΔGf° values:

[1ΔGf(CO2 (g)) + 2ΔGf(SO2 (g))] - [1ΔGf(CS2 (ℓ)) + 3ΔGf(O2 (g))]

[1(-394.38) + 2(-300.19)] - [1(65.27) + 3(0)] = -1060.03 kJ

-1,060.03 kJ (spontaneous)

From ΔG = ΔH - TΔS:

-1060.02 kJ (spontaneous)

EQUILIBRIUM CONSTANT, K (AT 298.15 K)

5.2426973554e+185

This process is favorable at 25°C.

Please mark as brainlest answer

Answer 2

Answer:

- 1075 kJ/mol

Explanation:

According to Hess's theory

ΔHfCS2(l) + ΔHreaction = ΔHfCO2(g) + 2 ΔHfSO2(g)

Therefore ,

ΔHreaction = ΔHfCO2(g) + 2 ΔHfSO2(g) - ΔHfCS2(l)

                   = (-393.5 + 2x-296.8 -87.9) kJ/mol

                   = - 1075 kJ/mol