Question

Can I get some help on this one?

For a particular reaction at 237.8 °C, ΔG = 339.93 kJ/mol, and ΔS = 519.96 J/(mol·K).

Calculate ΔG for this reaction at -65.7 °C.

Answer 1

\[\Delta G = \Delta H - T \Delta S \]

Answer 2

see that sign -Tdelta S. for starters when you increase the temperatrue. TDelta S gets bigger. this makes sense because as we increase temperature the particles move around faster and the randomness or Entropy increases. so the two go hand in hand.

Answer 3

ok makes sense

Answer 4

I'm wondering though if the enthalpy would be the same, that's probably the key here.
Because it's a measure of the energy of bonds broken - bonds formed. So I'm wondering if that's a molar quantity. Usually it's reported in kJ/mole

Answer 5

Entropy and temperature are directly related. but I'm wondering though if we can just

1. calculate the enthalpy for the reaction under the first temperature.

\[\Delta G + TDelta S = \Delta H_{1} \]

and then use the enthalpy along with the new temperature to calculate the new delta G.

\[\Delta G_{1} = \Delta H_{1}-T_{2}\Delta S \]


but for some reason I feel it should be more complicated than this.

Answer 6

My argument is that the amount of heat released enthalpy, is an intrinsic property that depends on the strength of the bonds themselves.

Answer 7

in this example we have to assume that the other values stay the same. It's true that enthalpy and entropy values usually do change with temperature, but we're assuming for simplicity's sake that they do not