If the change in free energy of a system is -899 Jmol^-1 and the entropy change is 10Jmol^-1k^-1 at 250C, calculate the enthalpy change.

darkypearl

11 years ago

where is2980gotn frm

Promix_Ose

1 year ago

∆G= -899
T= 25°C + 273= 298
∆S= 10
∆G = ∆H - T∆S

-899 = ∆H - (10 x 298)
-899 = ∆H - 2980
∆H = -899 + 2980 = + 2081 Jmol-1

Alka

10 years ago

T=25.... and T is absolute temperature which is measured in Kelvin. Therefore u multiply 25 by 273 which be equal to 298... den multiplied By 10(entropy change)

I.e.. 25×273=298

298×10=2980

Ifennaoffor

3 years ago

it's meant to be 25.0 celcius

emskodaddy

11 years ago

help me ask dem ooo

Florence Clifford

9 years ago

Can someone pls explain

Florence Clifford

9 years ago

I don't understand de solving

ICEBBEFC

9 years ago

YES HE IS CORRECT.AND IF YOU ARE STILL IN DOUGHT OF THIS QUESTION,SEARCH ON THIS TOPIC:STANDARD CONDITIONS FOR HEAT CHANGES.

Anisinclair

9 years ago

Is 25 constant at absolute temperature

Anisinclair

9 years ago

Is 25 constant at absolute temperature @Alka

Awedamoses

10 years ago

same here

DrUche

10 years ago

Hmmmmh...same question on my mind!