# How does an exothermic reaction change the entropy of the surroundings?

Jul 29, 2015

when we say that entropy always increases we refer to an isolated system and the entropy of the system and the environment, and for the formation of water, where entropy appears to decrease, we refer only to the system
However, in this case it is the entropy of the environment increases due to the heat yielded by Δ H = T Δ S
Samb Δ Δ = H / T = 484 kJ / 298 K = 1624 J / K, much higher than 88.9 J / K earned by the system
Δ = Δ Stot Samb Ssist + Δ = Δ H / T + Δ Ssist

Multiplying both sides by T
Stot Δ T = T + Δ -Δ Hsist Ssist
And calling Δ = G - T Δ Stot
We find the equation of Gibbs
Δ Δ G = H - T Δ S

Jul 29, 2015

An exothermic reaction increases the entropy of the surroundings.

#### Explanation:

The expression for the entropy change of the surroundings is:

$\Delta {S}_{s u r r} = - \frac{\Delta H}{T}$

For an exothermic reaction the sign of $\Delta H$ is -ve.

So $\Delta {S}_{s u r r}$ must be +ve.