Consider the reaction Ca(OH)2(aq) + 2HCl(aq)CaCl2(s) + 2H2O(l) for which H° = -3

Question

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.

(1) Calculate the entropy change of the UNIVERSE when 1.813 moles of Ca(OH)2(aq) react under standard conditions at 298.15 K. Suniverse = J/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'.

Explanation / Answer

1)

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

H° = -30.20 kJ

S° = 205.9 J/K at 298.15 K.

S° surrounings = - H° / T = 30.20 x 1000 / 298.15

                        = 101.29 J / K

S° universe = S° sys + S° surroundings

                    = 205.9 + 101.3

  S° universe = 307.2 J/K

2)

This reaction is product favoured.

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