Question #2f1c1

1 Answer
Ernest Z.
Mar 26, 2017

a) #ΔG^° = "-685.5 kJ"#; spontaneous.
b) #ΔG^° =color(white)(m)"-4.8 kJ"#; spontaneous.

Explanation:

Your essential formula is:

#color(blue)(bar(ul(|color(white)(a/a)ΔG^° = ΔH^° – TΔS^°color(white)(a/a)|)))" "#

a)

#"CH"_3"OH(l)" + "1.5O"_2("g") ⇌ "CO"_2"(g)" + "2H"_2"O(g)"#

#ΔH^° = "-638.4 kJ"#

#ΔS^° = "156.9 J·K"^"-1"#

#ΔG^° = "-638.4 kJ" - 298 color(red)(cancel(color(black)("K"))) × "0.1569 kJ"·color(red)(cancel(color(black)("K"^"-1"))) = "(-638.4 - 46.8) kJ" = "-685.2 kJ"#

#ΔG < 0#, so the reaction is spontaneous.

b)

#"2NO"_2"(g)" ⇌ "N"_2"O"_4"(g)"#

#ΔH^° = "-57.2 kJ"#

#ΔS^° = "-175.9 J·K"^"-1"#

#ΔG^° = "-57.2 kJ" + 298 color(red)(cancel(color(black)("K"))) × "0.1759 kJ"·color(red)(cancel(color(black)("K"^"-1"))) = "(-57.2 + 52.4) kJ" = "-4.8 kJ"#

#ΔG < 0#, so the reaction is spontaneous.

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