Temperature and total entropy change

The total entropy change (ΔS_total) of a system is given by:

ΔS_total = ΔS_system + ΔS_surroundings

Where:

• ΔS_system = entropy change of the system
• ΔS_surroundings = entropy change of the surroundings

ΔS_total is positive for all spontaneous changes. While ΔS_system doesn't vary much with temperature (unless there's a phase change), ΔS_surroundings is significantly affected.

The entropy change of the surroundings is calculated using:

ΔS_surroundings = $-\frac{\mathrm{\Delta }H}{T}$

Where:

• ΔH = enthalpy change of the reaction (J mol^-1)
• T = absolute temperature (K)

This equation shows that as temperature increases, the magnitude of ΔS_surroundings decreases. We can use this to determine if a reaction is spontaneous at a given temperature.

Worked example 1 - Determing reaction spontaneity at different temperatures

Consider the decomposition of magnesium carbonate:

MgCO_3(s) ➔ MgO_(s) + CO_2(g) ΔH = +118.5 kJ mol^-1, ΔS_system_ = +175.6 J K^-1 mol^-1

Determine the spontaneity of the reaction at 20°C and at 600°C.

Step 1: Conversion of °C into K

To convert from °C into K, add 273

20°C = 293 K

600°C = 773 K

Step 2: Calculate ΔS_total at 20°C

ΔS_surroundings = $-\frac{\mathrm{118,500}}{293}=-$404.4 K^-1 mol^-1

ΔS_total = +175.6 $-$ 404.4 = $-$228.8 J K^-1 mol^-1

Step 3: Determine spontaneity at 20°C

ΔS_total < 0, so reaction is not spontaneous

Step 4: Calculate ΔS_total at 600°C

ΔS_surroundings = $-\frac{\mathrm{118,500}}{773}=-$153.3 J K^-1 mol^-1

ΔS_total = +175.6 $-$ 153.3 = +22.3 J K^-1 mol^-1

Step 5: Determine spontaneity at 600°C

ΔS_total > 0, so reaction is spontaneous

This example demonstrates that the spontaneity of a reaction can change with temperature due to the impact on ΔS_surroundings.

The relationship between ΔStotal and the equilibrium constant (K)

For a reversible reaction at equilibrium, both the forward and reverse reactions occur spontaneously at equilibrium. This means ΔS_total must be positive in both directions.

Consider the equilibrium:

N₂O_4(g) ⇌ 2NO_2(g)

The graph below shows how it is possible for the reaction to be spontaneous in each direction.

This graph demonstrates that:

• At equilibrium, entropy is at a maximum. Moving away from equilibrium, whether toward pure reactants or products, decreases entropy.
• The change in entropy as you move from pure reactants or products to the equilibrium mixture is always positive. This makes both the forward and reverse reactions spontaneous.
• Moving from the equilibrium mixture toward pure reactants or products results in a negative entropy change, preventing the reaction from proceeding to completion in either direction.
• At equilibrium, ΔS_total(forward) = ΔS_total(reverse).

The relationship between ΔS_total and K is given by:

$\mathrm{\Delta }{S}_{\text{total}}=R\ln K$

Where:

• R = gas constant (8.31 J K^-1 mol^-1)
• K = equilibrium constant (K_c or K_p)

A large positive ΔS_total corresponds to a high K value.

Worked example 2 - Calculating K from ΔStotal

For the reaction: SO_2(g) + ½O_2(g) ⇌ SO_3(g), ΔS^⦵_total = 238.3 J K^-1 mol^-1

Calculate the value of the equilibrum constant (K). Give your answer to 3 significant figures.

Step 1: Equation

$\mathrm{\Delta }{S}_{\text{total}}=R\ln K$

Step 2: Rearrange equation

$\ln K=\frac{\mathrm{\Delta }{S}_{\text{total}}}{R}$

$K=e^{\frac{\mathrm{\Delta }{S}_{\text{total}}}{R}}$

Step 3: Substitution and correct evaluation

$K=e^{\frac{238.3}{8.31}}=2.84\times {10}^{12}$

This large K value suggests the equilibrium lies far to the right (products side).