very basic about ΔG

my first question in chemistry and understandably quite a dumb one too...

ΔG = ΔH - TΔS ( i understand this to be the def.)

and ΔS = q/T ( i understand even this to be some def.)

so ΔG = ΔH - q

where Δstands for change; G is gibbs free energy; S is entropy q is heat supplied and H is enthalpy
now in god's name can any one please explain what is the difference between q and ΔH i'll be very very thankful

8 Answers

1
skygirl ·

wow!! first question!!

1
Philip Calvert ·

first question not becoz i understand it well !!
you can't take in a more opposite sense sky.

now please can someone seriously help me out....

11
Mani Pal Singh ·

dekh bhai shayad tujhe pata ho
for a reversible isothermal process
∂S = 1/T X ∫dqrev
∂S =q/T

for reversible isothermal phase transformation at constant pressure
∂S = ∂H/T ( ∂H = qp)

1
Philip Calvert ·

you said "( ∂H = qp)"

toh kya ΔG =0 hoga... [7][7] yahi toh main pooch raha hun

11
Mani Pal Singh ·

the derivation of ΔG follows
look
G=H-TS
ΔG=ΔH-ΔTS
now look carefully
IF TEMPERATURE IS CONSTANT
thenΔG=ΔH-TΔS
and in the above explanation i have clearly mentioned the conditions req for ur assertion to be correct

11
rkrish ·

@philip
in your statement ∆S=q/T
1st of all q should be replaced by qrev as this statement is true only for a reversible process.

qrev : heat absorbed system
qrev=∆E+∆(PV)
∆H : heat change or heat of reaction(in a chemical process) at constant Pressure
∆H=∆E+P∆V

1
kIrTi kUmAr kAsAt ·

we know ∂H=qp(heat supplied at const. press.)
if qp=q then ∂G=0(EQUILIBRIUM)
if qp>q then ∂G>0(non spontaneous)
and ifqp-q<0 then ∂G<0(spontaneous)

1
Philip Calvert ·

[12] hmmmm

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