Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ?f H? = –286 kJ mol–1 ?

To calculate the entropy change in the surroundings when 1.00 mol of liquid water (H2O) is formed under standard conditions, we can use the relationship between enthalpy change and entropy change. The key concept here is that when a reaction occurs, the heat released or absorbed by the system affects the entropy of the surroundings. Let's break this down step by step.

Understanding the Relationship

The entropy change in the surroundings (ΔS_surroundings) can be calculated using the formula:

ΔS_surroundings = -ΔH / T

where:

• ΔH is the enthalpy change of the reaction (in joules),
• T is the temperature in Kelvin.

Given Data

From the problem, we know:

• The enthalpy change for the formation of 1.00 mol of H2O(l) is ΔH = -286 kJ/mol.
• Standard conditions imply a temperature of 298 K (25 °C).

Converting Units

First, we need to convert the enthalpy change from kilojoules to joules:

ΔH = -286 kJ/mol × 1000 J/kJ = -286,000 J/mol.

Calculating Entropy Change

Now, we can substitute the values into the formula:

ΔS_surroundings = -(-286,000 J/mol) / 298 K

ΔS_surroundings = 286,000 J/mol / 298 K

ΔS_surroundings ≈ 959.73 J/(mol·K).

Final Result

Thus, the entropy change in the surroundings when 1.00 mol of liquid water is formed under standard conditions is approximately:

ΔS_surroundings ≈ 959.73 J/(mol·K).

Conceptual Insight

This positive value for the entropy change in the surroundings indicates that the surroundings gain disorder as heat is released during the exothermic reaction of forming water. In essence, when the system (the formation of water) releases energy, it increases the randomness or disorder of the surrounding environment, which is reflected in the positive entropy change.