gibbs free energy

[velamj]

Hello all...im having problems understanding a problem in chemistry. Tbh I think this book may be wrong (1001 mcat chem questions examkrackers #438)

What is the standard Gibbs free energy of formation of water vapor at 25 degrees celsius if, for the reaction shown below under standard conditions, delta h = -484 KJ/MOL and delta S= -89 J/mol K ?

2H2 (g) + O2(g) ----> 2 H20(g)

a -457 kj/mol
b -395 kj/mol
c -229 kj/mol
d. water vapor does not form at 25 degrees celsius

I thought the answer was -457 but apparently its -229....

I know to convert 89j/mol k to .089 KJ/MOL K ....THIS conversion gives me -457....what am i doing wrong here???? i would appreciate any help i can get thanks

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[velamj]

ok i think i get it...the answer uing the formula gives -457 KJ/MOL......but since u have 2 moles of h20 its -457 kj/mol /2 mol = -229.....duh i get it im sorry

 

[Chocolatebear89]

wait why do you divide by 2?

 

[fizzgig]

they give you values for the reaction shown which yields 2mol of water.

the answers are given in kJ/mol. so you have 457kJ/2mol and 229kJ/1mol.

 

[vsl5]

but they gave delta H and delta S for the initial reaction per mole of product already.. So to calculate delta G for that reaction, you would have multiplied by 2 anyways, then divided by 2 at the end, giving you the initial value. Unless the OP accidently included per mol when he gave delta H and S values.

 

[fizzgig]

well monkey poop. now i'm confused. you're right, the units you get from the equatino itself are kJ/mol... and i can't imagine the book randomly giving values in J/#molesthatisn't1....

 

[Rabolisk]

The question asks what the Gibbs free energy of formation of water vapor is given the data for the reaction 2H2 + O2 -> 2H2O.

Standard deltaG of formation means 1 mol of H2O, but the reaction, and the data given, are for the reaction, which is 2 mols of H2O.