OpenStudy (anonymous):

A chemical reaction has a ΔHrxn of - 157 kj and a ΔSrxn of - 221 J/K. Is this reaction spontaneous at 525 K?

4 years ago
OpenStudy (anonymous):

Try using this equation: ΔG = ΔH - TΔS

4 years ago
OpenStudy (aaronq):

\(\Delta G=\Delta H - T\Delta S\) the reaction is spontaneous when \(\Delta G<0\)

4 years ago
OpenStudy (aaronq):

ahh you beat me to it lol

4 years ago
OpenStudy (anonymous):

^ I left out the spontaneity part lol

4 years ago
OpenStudy (aaronq):

team work !

4 years ago
OpenStudy (anonymous):

(y)

4 years ago
OpenStudy (anonymous):

thanks guys

4 years ago
OpenStudy (aaronq):

np dude !

4 years ago
OpenStudy (anonymous):

the next question says though, at what temperature is the reaction at equilibrium

4 years ago
OpenStudy (anonymous):

anyone know?

4 years ago
OpenStudy (anonymous):

I think, but not entirely sure, that's when G=0. So just solve for T algebraically.

4 years ago
OpenStudy (aaronq):

yep, he's right. the reaction is at eq. when \(\Delta G=0\)

4 years ago
OpenStudy (anonymous):

yes but how would set that up to solve

4 years ago
OpenStudy (anonymous):

It's just substituting the values back into ΔG=ΔH−TΔS. We know that ΔG=0, and that ΔH and ΔS are still the same as stated in the question. The only variable in the equation is T, and it's what you need to find out.

4 years ago