#CO (s) + S (s) + 2O_2 (g) -> CaSO_4 (s)# #DeltaH_(rxn)^@ = -888.3 " kJ/mol", DeltaS_(rxn)^@ = -118.0 "J/mol K"# At what temperature will this reaction be spontaneous?

Answer 1

#"T < 7528 K"#

At standard conditions

#"ΔG"^@ = Δ"H"^@ - "T"Δ"S"^@#

#"ΔG"^@ < 0# - Reaction is spontaneous #"ΔG"^@ = 0# - Reaction is at equilibrium #"ΔG"^@ > 0# - Reaction is non-spontaneous

For reaction to be at equilibrium

#0 = Δ"H"^@ - "T"Δ"S"^@#

#"T" = (Δ"H"^@)/(Δ"S"^@) = (-888.3 × 10^3\ "J/mol")/(-118.0\ "J/mol K") ≈ "7528 K"#

∴ Reaction is spontaneous at any temperature below #"7528 K"#

For more explanation on #"ΔG"# click here.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

#CO (s) + S (s) + 2O_2 (g) -&gt; CaSO_4 (s)# #DeltaH_(rxn)^@ = -888.3 " kJ/mol", DeltaS_(rxn)^@ = -118.0 "J/mol K"# At what temperature will this reaction be spontaneous?

#CO (s) + S (s) + 2O_2 (g) -> CaSO_4 (s)# #DeltaH_(rxn)^@ = -888.3 " kJ/mol", DeltaS_(rxn)^@ = -118.0 "J/mol K"# At what temperature will this reaction be spontaneous?