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When water is added to a mixture Na_2O_2(s) and S(s), a redox reaction occurs, as represented by the equation below.
2NaO_2(s) + 2 H_2O(l) rightarrow 4 NaOH (aq) + SO_2(aq) Delta H_298 degree = -610 kJ/mol_rxn; Delta S_298 degree = -73 J/(K mol_rxn)
Which of the following statements about the thermodynamic favorability of the reaction at 298 K is correct?
(A) It is thermodynamically unfavorable
(B) It is thermodynamically favorable and is driven by Delta S degree only.
(C) It is favorable and is driven by Delta H degree only.
(D) It is thermodynamically favorable and is driven by both Delta H degree and Delta S degree

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Answer:

(C) It is favorable and is driven by ΔH° only.

Explanation:

Let's consider the following balanced equation.

2 Na₂O₂ + S + 2 H₂O → 4 NaOH + SO₂

To determine whether it will be favorable or not at 298 K, we need to calculate the standard Gibbs free energy (ΔG°).

  • If ΔG° < 0, the reaction will be favorable.
  • If ΔG° > 0, the reaction will be unfavorable.

We can calculate ΔG° using the following expression.

ΔG° = ΔH° - T.ΔS°

As we can see from the expression above, the favorability will be driven if ΔH° < 0 and if ΔS° < 0. Since ΔH° = -610 kJ/mol and ΔS° = -73 J/K.mol, the favorability will be driven by ΔH°. Now, let's calculate the overall favorability.

ΔG° = ΔH° - T.ΔS°

ΔG° = -610 kJ/mol - 298 K.(-0.073 kJ/K.mol) = -588 kJ/mol

The reaction is favorable and is driven by ΔH° only.

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