Use the Gibbs free energy equation shown below to determine the temperature in degrees Celsius above which decomposition of KClO4 is spontaneous.
ΔG = ΔH – TΔS
2KClO4(s) → 2KClO3(s) + O2(g)
ΔH°f = 70.2 kJ
ΔS°f = 189.3 J/K
_______°C
2007-05-04 07:41:09 · 4 answers · asked by MEB 2 in Science & Mathematics ➔ Chemistry
The reaction is spontaneous when ∆G < 0
so:
0 < ∆H - T∆S ==> T > ∆H/(∆S*kJ/1000J)
∆H/(∆S*1000) = 70.2kJ/189J/K *kJ/1000J) = 371.43 K
That's above (371.43 - 273) = 98.42 ºC
2007-05-04 07:58:54 · answer #1 · answered by Dr Dave P 7 · 1⤊ 0⤋
Multiply the enthalpy value by 1,000 and divide it by the entropy value for your answer.
2007-05-04 14:45:08 · answer #2 · answered by Gervald F 7 · 1⤊ 0⤋
Do your own homework.
2007-05-04 14:48:17 · answer #3 · answered by Icarus 3 · 0⤊ 1⤋
its "i dunno" C do ur homework
2007-05-04 15:31:02 · answer #4 · answered by Dan M 1 · 0⤊ 1⤋