Answer :
Final answer:
The reaction NH₄Cl(aq) → NH₃(g) + HCl(aq) is favored under standard conditions at 345 K. The entropy change for the reaction of 1.72 moles of NH₄Cl(aq) at this temperature would be 0.079 kJ/K.
Explanation:
To determine whether the reaction NH₄Cl(aq) → NH₃(g) + HCl(aq) is favored under standard conditions at 345 K, we can use the Gibbs free energy change (△G°) and the equation △G° = △H° - T△S°.
Given that △G° = 59.1 kJ and △H° = 86.4 kJ, we can rearrange the equation to solve for △S°:
△S° = (△H° - △G°) / T
Substituting the values, we get:
△S° = (86.4 kJ - 59.1 kJ) / 345 K
△S° = 0.079 kJ/K
Therefore, the entropy change for the reaction of 1.72 moles of NH₄Cl(aq) at 345 K would be 0.079 kJ/K.
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Final answer:
The reaction NH₄Cl(aq) → NH₃(g) + HCl(aq) is favored under standard conditions at 345 K. The entropy change for the reaction of 1.72 moles of NH₄Cl(aq) at this temperature would be 0.079 kJ/K.
Explanation:
To determine whether the reaction NH₄Cl(aq) → NH₃(g) + HCl(aq) is favored under standard conditions at 345 K, we can use the Gibbs free energy change (△G°) and the equation △G° = △H° - T△S°.
Given that △G° = 59.1 kJ and △H° = 86.4 kJ, we can rearrange the equation to solve for △S°:
△S° = (△H° - △G°) / T
Substituting the values, we get:
△S° = (86.4 kJ - 59.1 kJ) / 345 K
△S° = 0.079 kJ/K
Therefore, the entropy change for the reaction of 1.72 moles of NH₄Cl(aq) at 345 K would be 0.079 kJ/K.
Learn more about favorability of a reaction and entropy change here:
https://brainly.com/question/31476138
#SPJ14