Consider the reaction Ca(OH) (ag)+2HCI(a)CaCl(s)+2H,00) for which ??\"--30.20 kJ

Question

Consider the reaction Ca(OH) (ag)+2HCI(a)CaCl(s)+2H,00) for which ??"--30.20 kJ and ?So 205.9 JK at 298.15 K. (1) Calculate the entropy change of the UNIVERSE when 2.475 moles of Ca(OH)2(aq) react under standard conditions at 298.15 K. ASuniverse 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

Answer 1) Entropy change of surrounding = - delta H/T = - (-30.2 kJ/298.15K) = 30200 J/298.15K = 101.29 J/K.

Entropy change of system = 205.9 J/K

Hence, Entropy change of universe = Entropy change of system + Entropy change of surroundings

Hence, delta S universe = 205.9 + 101.29 = 307.19 J/K

Ans 2) Since entropy is positive there is a disorderness or randomness in the system and hence the reaction is a spontaneous one . This reaction is reactant favoured under standard conditions as concentration of reactants is greater than products.

Ans 3 ) delta G = delta H - Tdelta S

if delta G is positive delta H is positive and delta S is never negative we know. So, if reaction is reactant favored it means it is ENTROPY FAVORED.

Ans 1    delta S surroundings = - delta H/T = -746.6 * 1000/298.15 = - 2504 J/K

delta S universe = - 2504 + 198 = - 2306 J/K

Ans 2 Reaction is product favored as concentration of products are greater than reactants.

Ans 3 The reaction is product favored and hence reaction is enthalpy favored.

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