For the gaseous reaction of carbon monoxide and chlorine to form phosgene, (COCl2), calculate each of the following.
(a) Calculate S° at 298 K
I got -136.76 J/K using delta S = sum of delta S of products minus sum of delta S of reactants which is the right answer.
But I can't figure out the second part:
(b) Assuming that S° and H° change little with temperature, calculate G° at 426 K.
In the standard thermodynamics values appendix it says the deltaH of COCl2 is -220 kJ/mol.
I am just not sure how to do this.
6 answers
First, this is at 426K and not 298 K; therefore, any number you obtain by dGo = dHo - TdSo will not agree with 220 kJ/mol. So you substitute the values you have for dHo and dSo along with 426K for T and calculate dGo. Remember dH is given in kJ/mol and dS in J/mol.
DeltaG=DeltaH-DeltaS*T
Where
DeltaH=-220kJ/mol
DeltaS=-136.76kJ/K
T=426K
and
Delta=????
Where
DeltaH=-220kJ/mol
DeltaS=-136.76kJ/K
T=426K
and
Delta=????
The key part of the equation states that ASSUME that DeltaS and DeltaH DOESN'T Change much, meaning that it is safe to plug and chug.
***To the questioner: change -136.76J/K to kJ, which should be -0.13676kJ/K
***To the questioner: change -136.76J/K to kJ, which should be -0.13676kJ/K
After reading Devron's reply I realize I read your initial question wrong. His response and mine agree; the only problem with mine is I thought you were quoting dG as -220 kJ/mol where you were quoting dH. Therefore, that part I wrote about "not agreeing with" make s no sense.
To the questioner: I apologize about the typos.
I meant to say solve for Delta*G and not just Delta. My post should have said the following:
DeltaG=DeltaH-DeltaS*T
Where
DeltaH=-220kJ/mol
DeltaS=-136.76J/K=-0.13676kJ/K
T=426K
and
DeltaG=???
I meant to say solve for Delta*G and not just Delta. My post should have said the following:
DeltaG=DeltaH-DeltaS*T
Where
DeltaH=-220kJ/mol
DeltaS=-136.76J/K=-0.13676kJ/K
T=426K
and
DeltaG=???
I initially made the same mistake, which confused me a little. I reread the question and saw that the questioner and I, both, may have initially overthought the question.
--Best
--Best