Question

Consider the reaction

2CO₂(g) + 5H₂(g)C₂H₂(g) + 4H₂O(g)

for which H° = 46.50 kJ and S° = -124.8 J/K at 298.15 K.

(1) Calculate the entropy change of the UNIVERSE when 1.850 moles of CO₂(g) react under standard conditions at 298.15 K.

S_universe =   J/K

(2) Is this reaction reactant or product favored under standard conditions?

_________reactantproduct

(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'.

_________enthalpyentropybothreactant favored

Submit Answer

Answer 1

2CO2(g) + 5H2(g)C2H2(g) + 4H2O(g)

given

ΔHrxn = 46.50 KJ
ΔSrxn = -124.8 J/k

This values for 2 mols CO2 is used
when 1.850 mol CO2 is used,

ΔHrxn = 46.50 KJ/2 mol* 1.850 mol = 43.0125 x10^3 J

ΔSrxn = -124.8 J/k/2 mol * 1.850 mol = -115.44 J/K

ΔSsurrounding =-ΔHrxn/T
=-43.0125 x10^3 J J/298.15 K= -144.26 J/K

ΔSuniverse = ΔSsurrounding + ΔSrxn
= -144.26 J/K + -115.44 J/K
ΔSuniverse = -259.7 j/k
*******************

ΔSuniverse is negative and ΔHrxn positive Hence
This reaction is reactant favored under standard conditions
enthalpy and entropy both reactant favored