Consider the reaction CH4(R)+202) CO2(R)+2H20() for which ??\'--802.3 kJ and ? S

Question

Consider the reaction CH4(R)+202) CO2(R)+2H20() for which ??'--802.3 kJ and ? S.--5.200 JK at 298.15 ?. (1) Calculate the entropy change of the UNIVERSE when 1.692 moles of CH4(g) react under standard conditions at 298.15 ?. ? 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

Given:

CH4 (g) + 2O2 (g) -----------------> CO2 (g) + 2H2O (g)

?S0 = -5.200 J/K , ?H0 = -802.3 KJ,

to find: ?S Universe

?S Universe = ?Ssystem + ?Ssurroundings

Reaction itself is the system here, so ?Ssystem = -5.200 J/K

?Ssurrounding = -?H0/T = -(-802300 J / 298.15 K = + 2690.9 J/K

?S Universe = - 5.200 J/K + (2690.9 J/K ) = 2685.7 J/K

2. Under these conditions Since ?S Universe is positive, the reaction would be product favored.

3. The reaction would be enthalpy favored as the enthalpy is a large negative number. Both delat H and delta S are negative.

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