15.2 Entropy and Spontaneity

What is a spontaneous reaction?

A spontaneous reaction is one that can happen under the conditions being studied.

It may happen immediately, or it may be hampered by a very high Activation Energy.

Exothermic reactions are often spontaneous under normal conditions but some endothermic changes happen spontaneously too.

So it's not all to do with Enthalpy Change.

Physicists get annoyed when people say that Entropy is a measure of disorder because it's more complicated than that.

But, the more disorder a system the higher its entropy will be.

So reactions that create more disorder will have Positive entropy changes.

QUESTION: Explain why the change in entropy is so much bigger during boiling than melting.

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QUESTION: Predict whether the following reactions will have positive or negative entropy changes.

Thankfully, we don't need to know how Standard Entropy Changes are calculated - just how to use them

ENTROPY CHANGE (∆S) = Total Entropy of Products - Total Entropy of Reactants

QUESTION: Use the data above to calculate the standard entropy change for:

H_2(g) + O_2(g) -> 2H₂O _(g)

N_2(g) + 3H_2(g) -> 2NH_{3 (g)}

H_2(g) + Cl_2(g) -> 2HCl _(g)

Again, we don't need to really understand what Gibbs Free Energy is so long as we know how to use it.

The equation:

∆G = ∆H - T∆S

is given in the data book, though you should probably learn it anyway.

Reactions are spontaneous when ∆G = 0 or ∆G = negative

QUESTION:Complete the table below

Calculating ∆G reaction from Standard Free Energies

Standard Free Energy Changes are calculated like Entropy changes

FREE ENERGY CHANGE (∆G) = Total Free Energy of Products - Total Free Energy of Reactants

QUESTION: Use the data above to calculate the standard entropy change for:

H_2(g) + O_2(g) -> 2H₂O _(g)

N_2(g) + 3H_2(g) -> 2NH_{3 (g)}

2CO_(g) + O_2(g) -> 2CO_{2 (g)}

Use the equation ∆G = ∆H - T∆S and the data above to show whether the following reactions are feasible at 400 K. Make sure you convert J/K/mol to kJ/K/mol

H_2(g) + O_2(g) -> 2H₂O _(g) ∆H_R = -572 kJ/mol

N_2(g) + 3H_2(g) -> 2NH_{3 (g)} ∆H_R = -186 kJ/mol

Reactions are first spontaneous when ∆G = 0

Rearranging the Gibbs equation then gives us ∆H = T∆S

And T= ∆H /∆S

Find the first temperature at which these reactions become feasible

H_2(g) + O_2(g) -> 2H₂O _(g) ∆H_R = -572 kJ/mol

N_2(g) + 3H_2(g) -> 2NH_{3 (g)} ∆H_R = -186 kJ/mol