Question

What is the DG under standard temperature for the following reaction between?
NH3(s) -> NH3(l)
Given:
NH3(s): DHf = -46.11 kJ; DSf = 192.45 J/K
NH3(l): DHf = -80.29 kJ; DSf = 111.3 J/K

Answer 1

find the DH of the reaction, find the DS of the reaction, then plug into DG=DH-TDS

Answer 2

would the DH be -34.18?

Answer 3

I am sorry but I can't help with this one sorry/ :(

Answer 4

DH, DS, and DG are always products - reactants, so is that the right order?

Answer 5

@JFraser I think it is
(-80.29 kJ)-(-46.11)= -34.18

Answer 6

so that's the DH, do the same thing to find the DS

Answer 7

I have that too, DS=-81.15

Answer 8

What I don't know is how to set up or plug in the info into the equation
DG=DH-TDS

Answer 9

you just found DH and DS, what's the temperature?

Answer 10

well.. It says to use the standard temperature, so that would be
273.15 K..?

Answer 11

correct, so plug all 3 pieces into \[\Delta G = \Delta H - T * \Delta S\] and be sure to convert the KJ and J into the same unit

Answer 12

okay, Thank you @JFraser