Entropy of a chemical reaction

1. Mar 18, 2016

erisedk

1. The problem statement, all variables and given/known data
The entropy change of surrounding is greater than that of the system in a/an
(A) exothermic process
(B) endothermic process
(C) both (A) and (B) are correct
(D) none of these are correct

2. Relevant equations

3. The attempt at a solution
ΔStot = ΔSsys + ΔSsurr
For a spontaneous process, ΔStot > 0
For an exothermic process, ΔH < 0 and for an endothermic process ΔH > 0

2. Mar 18, 2016

Ygggdrasil

Try looking up a formula that relates $\Delta S_{surr}$ to $\Delta H$. Remember that for chemical reactions performed at constant pressure, the change in enthalpy tells you the amount of heat transferred between the system and surroundings.

3. Mar 18, 2016

erisedk

ΔG = ∆Hsys-T∆Ssys
At equilibrium ∆G = 0
So ∆Ssys = ∆Hsys/T
For exothermic reactions, ∆Hsys < 0 so ∆Ssys < 0
For ∆Ssys + ∆Ssurr > 0
∆Ssurr > 0
Since a positive number is always greater than a negative number, ∆Ssurr > ∆Ssys
So A is correct.
For endothermic reactions, ∆Ssys > 0 as ∆Hsys > 0
Now for ∆Ssys + ∆Ssurr > 0
We can have ∆Ssurr as
1. positive and larger than system entropy
2. Positive and smaller than system entropy
3. Negative but smaller in magnitude than system entropy.

So which of these would be appropriate?

4. Mar 19, 2016

haruspex

The question asks which entropy change is the greater. I would interpret that in terms of magnitude.

5. Mar 19, 2016

Ygggdrasil

A useful formula here is to remember that the entropy change of the surroundings depends on the amount of heat transferred to/from the system: $\Delta S_{surr} = - \Delta H/T$

With this in mind (and the fact that for any spontaneous process $\Delta G < 0$), how do the magnitudes of $\Delta S_{sys}$ and $\Delta S_{surr}$ compare in the two cases?