Consider the reaction 2C2H6(g) +702(g)4CO2(g) 6H2O(g) for which Ho--2.855x 103 k

Question

Consider the reaction 2C2H6(g) +702(g)4CO2(g) 6H2O(g) for which Ho--2.855x 103 kJ and S= 92.70 JK at 298.15 K. (1) Calculate the entropy change of the UNIVERSE when 1.836 moles of C2H(g react under standard conditions at 298.15 K J/K universe (2) Is this reaction reactant or product favored under standard conditions? (3) If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the reaction is reactant favored, choose 'reactant favored'

Explanation / Answer

1) We have H0 = -2.855*103 kJ, S0 = 92.70 J/K and T = 298.15 K.

We have the enthalpy change, H0 for 2 moles of C2H6. H0 and S0 are extensive propertIES, i.e, depends on the amount of the reactants and hence, for 1.836 mole of C2H6, we must have

H0’ = (H0)*(1.836 mole C2H6/2 mole C2H6) = (-2.855*103 kJ)*(1.836 mole C2H6/2 mole C2H6) = -2620.89 kJ.

S0’ = (S0)*( 1.836 mole C2H6/2 mole C2H6) = (92.70 J/K)*( 1.836 mole C2H6/2 mole C2H6) = 85.0986 J/K.

S0’surr = -H0’/T = -(-2620.89 kJ)*(1000 J/1kJ)/(298.15 K) = 8790.5081 J/K.

Suniv = S0’ + S0’surr = (85.0986 J/K) + (8790.5081 J/K) = 8875.6067 J/K = (8875.6067 J/K)*(1 kJ/1000 J) = 8.8756 kJ/K (ans).

2) Calculate the free energy change for the reaction as

G0 = H0 – T*S0 = -(2.855*103 kJ) – (298.15 K)*(92.70 J/K)

= (-2.855*103 kJ) – (27638.505 J) = (-2.855*103 kJ) – (27638.505 J)*(1 kJ/1000 J)

= (-2.855*103 kJ) – (27.638505 kJ)

= -2882.638505 kJ -2.883*103 kJ (ans).

The free energy change for the reaction is negative, i.e, G0 < 0 and hence, the reaction is spontaneous in the forward direction, i.e, the reaction favors the products (ans).

3) The free energy change for the reaction favors the products as seen above. The reaction has a positive entropy change and hence, the reaction is entropically favored. The enthalpy change for the reaction is negative, i.e, the reaction produces energy and is exothermic. Consequently, the reaction is enthalpy favored as well. Therefore, the reaction is both enthalpy as well as entropy favored and hence, product favored (ans).

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