If ΔH = -70.0 kJ and ΔS = -0.300 kJ/K , the reaction is spontaneous below a certain temperature. Calculate that temperature.?
回答 (3)
2020-05-12 8:47 am
ΔG = ΔH - TΔS
When the reaction is spontaneous, ΔG < 0
ΔH - TΔS < 0
(-70.0) - T(-0.300) < 0
-70.0 + 0.300T < 0
0.300T < 70.0
T < 70.0/0.300
T < 233 K
The reaction is spontaneous below 233 K.
When the reaction is spontaneous, ΔG < 0
ΔH - TΔS < 0
(-70.0) - T(-0.300) < 0
-70.0 + 0.300T < 0
0.300T < 70.0
T < 70.0/0.300
T < 233 K
The reaction is spontaneous below 233 K.
2020-05-12 8:41 am
delta G = delta H - T delta S
d G = -70 - T(-0.3) = 0
T = -70/-.3 = 233.333 K rund for sig figs
d G = -70 - T(-0.3) = 0
T = -70/-.3 = 233.333 K rund for sig figs
2020-05-12 8:40 am
start with..
.. dG = dH - TdS
. .dG < 0, the rxn is spontaneousso to solve this.. (1) set dG = 0.. solve for T... (2) verify Temps less that (1) give dG < 0
.. dG = dH - TdS
. .dG < 0, the rxn is spontaneousso to solve this.. (1) set dG = 0.. solve for T... (2) verify Temps less that (1) give dG < 0
收錄日期: 2021-04-18 18:32:32
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