Question

(1) Calculate the entropy change of the UNIVERSE when 1.734 moles of Ca(OH)2(aq) react under standard...

(1) Calculate the entropy change of the UNIVERSE when 1.734 moles of Ca(OH)2(aq) react under standard conditions at 298.15 K.
Consider the reaction

Ca(OH)2(aq) + 2HCl(aq)CaCl2(s) + 2H2O(l)

for which H° = -30.20 kJ and S° = 205.9 J/K at 298.15 K.

(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'.

Homework Answers

Answer #1

1.

deltaSsys = 205.9 J / K

deltaHsys = - 30.20 kJ

T = 298.15 K

deltaSsurr = - deltaHsys / T = - ( - 30.20 ) / 298.15 = 0.1013 kJ / K = 101.3 J / K

Therefore,

deltaSuniv = deltaSsys + deltaSsurr = 205.9 + 101.3 = 307.2 J / K .mol

The, entropy of universe of 1.734 mol = 307.2 x 1.734 = 532.7 J / K

2.

Since, Entropy of universe > 0, the reaction is product favoured.

3.

The reaction is favoured by both enthalpy and entropy.

Because, enthalpy is negative and entropy is positive. Both are favourable for the formation of product i.e for spontaneous reaction.

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