Calculating entropy changes for reactions

During a reaction, the entropy change of the system (ΔS_system) can be calculated using the formula:

ΔS_system = S_products $-$ S_reactants

Where:

• S_products is the sum of the standard entropies of all products.
• S_reactants is the sum of the standard entropies of all reactants.

Standard entropy (S^⦵) is the entropy of 1 mole of a substance under standard conditions (100 kPa pressure and 298 K temperature). Its units are J K^-1 mol^-1.

The steps for calculating ΔS_system are:

1. Find the sum of standard entropies for the products (ΣS^⦵(products)).
2. Find the sum of standard entropies for the reactants (ΣS^⦵(reactants)).
3. Subtract the sum of reactant entropies from the sum of product entropies.

If the balanced equation has coefficients other than 1, multiply the standard entropy of each substance by its coefficient.

Worked example 1 - Calculating entropy change for a reaction

Calculate the entropy change for the reaction where hydrogen chloride reacts with ammonia to produce ammonium chloride at standard conditions:

HCl_(g) + NH_3(g) ➔ NH₄Cl_(s)

Given:

• S^⦵[HCl_(g)] = 186.9 J K^-1 mol^-1
• S^⦵[NH_3(g)] = 192.8 J K^-1 mol^-1
• S^⦵[NH₄Cl_(s)] = 94.6 J K^-1 mol^-1

Step 1: Calculate the entropy of the products

S_products = S^⦵[NH₄Cl] = 94.6 J K^-1 mol^-1

Step 2: Calculate the entropy of the reactants

S_reactants = S^⦵[HCl] + S^⦵[NH₃]

= 186.9 + 192.8 = 379.7 J K^-1 mol^-1

Step 3: Equation

ΔS_system = S_products $-$ S_reactants

Step 4: Substitution and correct evaluation

ΔS_system = 94.6 $-$ 379.7 = $-$285.1 J K^-1 mol^-1

The negative ΔS_system indicates a decrease in entropy, reflecting the increased order as two gases form a solid.

Total entropy change

To determine if a reaction is feasible, we need to consider the total entropy change of both the system and its surroundings:

ΔS^⦵_total = ΔS^⦵_system + ΔS^⦵_surroundings

The entropy change of the surroundings can be calculated using the following equation:

ΔS^⦵_surroundings = $-\frac{\mathrm{\Delta }{H}^{\ominus }}{T}$

Where:

• ΔH^⦵ = standard enthalpy change of the reaction (J mol^-1).
• T is the temperature (K).

ΔS_total_ will be positive, indicating a spontaneous process, when:

• Both ΔS_system_ and ΔS_surroundings_ are positive.
• ΔS_surroundings_ is positive and larger in magnitude than a negative ΔS_system_.
• ΔS_system_ is positive and larger in magnitude than a negative ΔS_surroundings_.

Worked example 2 - Calculating total entropy change

Calculate the total entropy change, in J K^-1 mol^-1, for the reaction of ammonia with hydrogen chloride:

NH_3(g) + HCl_(g) ➔ NH₄Cl_(s)

Given:

• ΔH^⦵ = $-$315 kJ mol^-1 (at 298 K)
• ΔS^⦵_system = $-$285.1 J K^-1 mol^-1 (calculated in worked example 1)

Step 1: Conversion of kJ mol^-1 into J mol^-1

To convert from kJ mol^-1 into J mol^-1, multiply by 1,000

$-$315 kJ mol^-1 = $-$315,000 J mol^-1

Step 2: Calculate ΔS^⦵_surroundings

ΔS^⦵_surroundings = $-\frac{\mathrm{\Delta }{H}^{\ominus }}{T}=-\frac{\mathrm{315,000}}{298}$ = +1,057 J K^-1 mol^-1

Step 3: Calculate ΔS^⦵_total

ΔS^⦵_total = ΔS^⦵_system + ΔS^⦵_surroundings = $-$285.1 + 1,057 = +772 J K^-1 mol^-1

The positive total entropy change suggests that this reaction is likely to be feasible under standard conditions.

Worked example 3 - Calculating total entropy change

Calculate the total entropy change, in J K^-1 mol^-1, for the dissolution of ammonium nitrate crystals in water under standard conditions:

NH_4_NO_3(s)_ ➔ NH_4_^+^(aq)** + NO_3_^-^(aq)**

Given:

• ΔH = +25.70 kJ mol^-1^ (at 298 K)
• S^⦵[NH_4_NO_3_] = 151.1 J K^-1^ mol^-1^
• S^⦵[NH_4_^+^(aq)] = 113.4 J K^-1^ mol^-1^
• S^⦵[NO_3_^-^(aq)] = 146.4 J K^-1^ mol^-1^

Step 1: Conversion of kJ mol^-1 into J mol^-1

To convert from kJ mol^-1^ into J mol^-1^, multiply by 1,000

25.70 kJ mol^-1 = 25,700 J mol^-1

Step 2: Calculate ΔS^⦵_surroundings

ΔS^⦵_surroundings = $-\frac{\mathrm{\Delta }{H}^{\ominus }}{T}=-\frac{\mathrm{25,700}}{298}=-$86.24 J K^-1 mol^-1

Step 3: Calculate ΔS^⦵_system

ΔS^⦵_system = ΔS^⦵_products $-$ ΔS^⦵_reactants = 113.4 + 146.4 $-$ 151.1 = +108.7 J K^-1 mol^-1

Step 4: Calculate ΔS^⦵_total

ΔS^⦵_total = ΔS^⦵_system + ΔS^⦵_surroundings = 108.7 $-$ 86.24 = +22.5 J K^-1 mol^-1

Worked example 4 - Calculating total entropy change

Calculate the total entropy change, in J K^-1 mol^-1, for the reaction between barium hydroxide and ammonium chloride:

Ba(OH)2**•8H_2_O_(s)_ + 2NH_4_Cl_(s)_ ➔ BaCl_2(s)_ + 10H_2_O_(l)_ + 2NH_3(g)_

Given:

• ΔH = +164.0 kJ mol^-1^ (at 298 K)
• S^⦵[Ba(OH)2**•8H_2_O] = 427.0 J K^-1^ mol^-1^
• S^⦵[NH_4_Cl] = 94.6 J K^-1^ mol^-1^
• S^⦵[BaCl_2_] = 123.7 J K^-1^ mol^-1^
• S^⦵[H_2_O] = 69.9 J K^-1^ mol^-1^
• S^⦵[NH_3_] = 192.3 J K^-1^ mol^-1^

Step 1: Conversion of kJ mol^-1 into J mol^-1

To convert from kJ mol^-1^ into J mol^-1^, multiply by 1,000

164.0 kJ mol^-1 = 164,000 J mol^-1

Step 2: Calculate ΔS^⦵_surroundings

ΔS^⦵_surroundings = $-\frac{\mathrm{\Delta }{H}^{\ominus }}{T}=-\frac{\mathrm{164,000}}{298}=-$550.3 J K^-1 mol^-1

Step 3: Calculate ΔS^⦵_system

ΔS^⦵_system = ΔS^⦵_products $-$ ΔS^⦵_reactants **

ΔS^⦵_system *= (*123.7 + (10 x 69.9) + (2 x 192.3)) $-$(427.0 + (2 x 94.6) = +591.1 J K^-1 mol^-1

Step 4: Calculate ΔS^⦵_total

ΔS^⦵_total = ΔS^⦵_system + ΔS^⦵_surroundings = 591.1 $-$ 550.3 = +40.8 J K^-1 mol^-1

The positive ΔS_total_ indicates that the reaction is spontaneous, despite being endothermic. This spontaneity is driven by the significant increase in entropy of the system, which arises from the formation of liquid water and gaseous ammonia as products.

Thermodynamic vs. kinetic stability

While total entropy determines if a reaction is thermodynamically feasible, it doesn't guarantee that the reaction will occur immediately. This is where the concept of kinetic stability becomes important.

Consider the combustion of methane:

CH_4(g) + 2O_2(g) ➔ CO_2(g) + 2H₂O_(l)

At 298 K:

ΔH^⦵ = $-$890.3 kJ mol⁻¹

ΔS_total = +2.95 kJ mol⁻¹

Despite having a positive ΔS_total, the combustion of methane doesn't occur spontaneously at 298 K because:

• It requires an ignition source to overcome its high activation energy.
• This makes the methane-oxygen mixture kinetically stable, even though it's thermodynamically unstable.

The impact of temperature on entropy

Temperature plays a significant role in determining changes in entropy.

When heat is added to a system:

• At lower temperatures, the increase in entropy is more substantial because molecules gain greater freedom of movement.
• At higher temperatures, the same amount of heat results in a smaller increase in entropy, as the molecules are already moving vigorously.

This dependence on temperature is crucial for understanding various natural phenomena, especially phase changes. For example, the concepts of entropy can explain why water freezes spontaneously at temperatures below its melting point.

Entropy changes during water freezing

Consider the process: H₂O_(l) ➔ H₂O_(s).

The following values are known:

• ΔH^⦵ = $-$6.01 kJ mol^-1
• S^⦵(H₂O_(l)) = 69.9 J K^-1 mol^-1
• S^⦵ (H₂O_(s)) = 47.9 J K^-1 mol^-1

For water to freeze spontaneously, the magnitude of ΔS_surroundings must be greater than the magnitude of ΔS_system, resulting in a positive ΔS_total.

Key observations:

• ΔS_system_ remains constant because it depends on the inherent properties of water and ice.
• ΔS_surroundings_ increases at lower temperatures, as the same amount of heat released causes a larger entropy change in colder surroundings.
• At $-$5°C, the increase in ΔS_surroundings_ is sufficient to make ΔS_total_ positive, allowing freezing to occur spontaneously.

This example shows why we use freezers to make ice. The lower temperature increases ΔS_surroundings_, which compensates for the decrease in entropy that occurs when water freezes, making the process thermodynamically spontaneous.