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### ΔS=q/t

Posted: Wed Feb 13, 2019 10:04 am
When looking at the equation ΔS=q/t, could we say that ΔS=nΔH/T as well?

### Re: ΔS=q/t

Posted: Wed Feb 13, 2019 10:06 am
ΔS=ΔH/T at constant pressure and temperature.

### Re: ΔS=q/t

Posted: Wed Feb 13, 2019 4:44 pm
What are the requirements to be able to use q/T? Just constant temp?

### Re: ΔS=q/t

Posted: Wed Feb 13, 2019 8:33 pm
you can use it when t is constant and you can use deltaS=DeltaH/T when pressure is constant

### Re: ΔS=q/t

Posted: Fri Feb 15, 2019 2:17 pm
This can only be said when pressure is constant since q(at constant pressure)= delta H

### Re: ΔS=q/t

Posted: Fri Feb 15, 2019 2:28 pm
Yes, you can use this when pressure is constant since the change in enthalpy is equal to q when pressure is constant.

### Re: ΔS=q/t

Posted: Fri Feb 15, 2019 2:36 pm
Why can you only calculate delta S, when the q for the reaction is reversible?

### Re: ΔS=q/t

Posted: Sat Feb 16, 2019 5:45 am
Soumya Ravichandran 4H wrote:Why can you only calculate delta S, when the q for the reaction is reversible?

The reaction does not have to be reversible in order to calculate delta S in this way. They just use q rev as a convention, but since delta S is a state function, the pathway doesn’t matter (therefore, it doesn’t matter if it’s reversible or reversible). The only thing we look at is final - initial. Which would be the same for irreversible and reversible.

### Re: ΔS=q/t

Posted: Sat Feb 16, 2019 9:54 pm
when pressure and temperature are constant, $\Delta S = \Delta Hsys/T$