Consider the reaction 4HCl(g) + O2(g)2H2O(g) + 2Cl2(g) for which H° = -114.4 kJ
ID: 1037376 • Letter: C
Question
Consider the reaction 4HCl(g) + O2(g)2H2O(g) + 2Cl2(g) for which H° = -114.4 kJ and S° = -128.9 J/K at 298.15 K.
(1) Calculate the entropy change of the UNIVERSE when 2.084 moles of HCl(g) 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) 4HCl(g) + O2(g) -----> 2H2O(g) + 2Cl2(g)
H° = -114.4 kJ and S° = -128.9 J/K at 298.15 K.
DG0 = DH0-TDS0
= (-114.4)-(298.15*-128.9*10^-3)
= -75.97 kj/mol
DG0sys = - TDSuniv
DSuniv = -DG0/T
= - (-75.97*10^3)/298.15
DSuniv = 254.8 j/k
for 2.084 moles HCl , DSuniv = 254.8*2.084/4 = 132.75 j/k
answer: 132.75 j/k
2) as DG0 = - ve, it is product favoured.
3) DG0 is -ve , due to -Ve sign of DH.so that, it is enthalpy favored.
answer; enthalpy favored
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