Consider the reaction S(s,rhombic) + 2CO(g) SO2(g) + 2C(s,graphite) for which Ho

Question

Consider the reaction S(s,rhombic) + 2CO(g) SO2(g) + 2C(s,graphite) for which Ho = -75.80 kJ and So = -167.6 J/K at 298.15 K.

a) Calculate the entropy change of the UNIVERSE when 1.685 moles of S(s,rhombic) react under standard conditions at 298.15 K. Suniverse = J/K

b) Is this reaction reactant or product favored under standard conditions?

c) 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

(a) delta H0 for 1.685 mol of Srhomic = - 75.80 * 1.685 = - 127.7 kJ

delta S0 for 1..685 mol of Srhomic = - 167.6 * 1.685 = - 282.4 J = - 0.2824 kJ/K

Delta G0sys= delta H0sys - T delta S0sys = - 127.7 - (298.15)( - 0.2824) = - 43.50 kJ

And it is known that,

Delta G0sys= delta H0sys - T delta S0sys

Delta G0sys= - delta H0 surr - T delta S0sys

Delta G0sys= - T delta S0surr - T delta S0sys

Delta G0sys= - T (delta S0surr + delta S0sys )

Delta G0sys= - T delta S0univ

delta S0univ = - Delta G0sys / T = - ( - 43.50) / 298.15 = 14.6 kJ/K = 0.0146 J/K

(b) It is product favoured because Delta G0 = -ve

(c) Enthalpy favoured.

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