Question

1- Calculate ΔG o for the following reaction at 25°C. You will have to look up the thermodynamic data. 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l)

2- A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −320.1 kJ/mol and ΔS = −99.00 J/K ·mol,determine the temperature (in °C)below which the reaction is spontaneous.

Answer 1

1)

Given:

Gof(C2H6(g)) = -32.82 KJ/mol

Gof(O2(g)) = 0.0 KJ/mol

Gof(CO2(g)) = -394.359 KJ/mol

Gof(H2O(l)) = -237.129 KJ/mol

Balanced chemical equation is:

2 C2H6(g) + 7 O2(g) ---> 4 CO2(g) + 6 H2O(l)

ΔGo rxn = 4*Gof(CO2(g)) + 6*Gof(H2O(l)) - 2*Gof( C2H6(g)) - 7*Gof(O2(g))

ΔGo rxn = 4*(-394.359) + 6*(-237.129) - 2*(-32.82) - 7*(0.0)

ΔGo rxn = -2934.57 KJ

Answer: -2934.6 KJ

2)

ΔG = 0.0 KJ/mol

ΔH = -320.1 KJ/mol

ΔS = -99 J/mol.K

= -0.099 KJ/mol.K

use:

ΔG = ΔH - T*ΔS

0.0 = -320.1 - T *-0.099

T = 3233 K

= (3233-273) oC

= 2960 oC

Answer: 2960 oC