Question

For the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(1)\), the value of \(\Delta \mathrm{H}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(\Delta \mathrm{S}=0.163\) \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\). The free energy change at \(300 \mathrm{~K}\). for the reaction, is: (a) \(-289.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(437.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-334.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-291.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Short answer

The free energy change is closest to -289.6 kJ/mol, so the answer is (a).

Step-by-step solution

Identify the Formula

We need the formula for calculating the Gibbs free energy change \( \Delta G \). The formula is given by:\[ \Delta G = \Delta H - T \Delta S \] where \( \Delta G \) is the change in free energy, \( \Delta H \) is the enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the entropy change.

Substitute Known Values

We are given \( \Delta H = -285.8 \ \mathrm{kJ \ mol}^{-1} \), \( \Delta S = 0.163 \ \mathrm{J \ K}^{-1} \mathrm{~mol}^{-1} \), and \( T = 300 \ \mathrm{K} \). First, we need to convert \( \Delta S \) from \( \mathrm{J} \) to \( \mathrm{kJ} \) to match the units of \( \Delta H \). \[ \Delta S = 0.163 \ \mathrm{J \ K}^{-1} \mathrm{~mol}^{-1} = 0.000163 \ \mathrm{kJ \ K}^{-1} \mathrm{~mol}^{-1} \]Now substitute these values into the formula:\[\Delta G = -285.8 - (300 \times 0.000163)\]

Perform the Calculation

Calculate the term \( T \Delta S \):\[ 300 \times 0.000163 = 0.0489 \ \mathrm{kJ \ mol}^{-1} \]Now compute \( \Delta G \):\[ \Delta G = -285.8 - 0.0489 = -285.8489 \ \mathrm{kJ \ mol}^{-1} \]

Select the Closest Answer

The calculated \( \Delta G \) is \(-285.8489 \ \mathrm{kJ \ mol}^{-1} \), which is closest to option (a) \(-289.6 \ \mathrm{kJ \ mol}^{-1}\). Therefore, the correct answer is:(a) \(-289.6 \ \mathrm{kJ \ mol}^{-1}\).

Key concepts

Enthalpy Change

Understanding enthalpy change is crucial in thermodynamics. Enthalpy change, denoted as \(\Delta H\), represents the heat absorbed or released during a chemical reaction at constant pressure. It helps in determining whether a reaction is exothermic or endothermic.
For exothermic reactions, \(\Delta H\) is negative as the system releases heat to the surroundings. For the given reaction of hydrogen gas with oxygen to form water, \(\Delta H\) is \(-285.8 \ \, \mathrm{kJ \ mol}^{-1}\), indicating an exothermic process. This means energy is released, making the environment around the reaction warm.
Knowing the sign and magnitude of \(\Delta H\) helps in predicting how the reaction will affect its surroundings. Positive values denote that heat is absorbed, categorizing it as an endothermic reaction. Understanding enthalpy provides insight into energy changes in chemical systems.

Entropy Change

Entropy change, represented as \(\Delta S\), is a measure of disorder or randomness in a system. In thermodynamics, understanding entropy helps predict the spontaneity of a process.
In our exercise, the entropy change is \(0.163 \, \mathrm{J \ K}^{-1} \mathrm{~mol}^{-1}\). This indicates a slight increase in disorder as the reaction proceeds from reactants to products. Entropy is a crucial factor when calculating Gibbs free energy.

• Positive \(\Delta S\) implies that the disorder increases, which tends to favor spontaneous reactions.
• Negative \(\Delta S\) implies a decrease in disorder, usually making the process non-spontaneous unless driven by other factors like enthalpy change.

Understanding entropy is essential for predicting how changes in temperature and pressure will influence the favorability of reactions.

Thermodynamics in Chemistry

Thermodynamics in chemistry involves studying energy changes and the physical properties of matter during chemical reactions and physical transformations.
The Gibbs free energy, \(\Delta G\), combines enthalpy and entropy to predict the spontaneity of a reaction. It is calculated with the formula: \[ \Delta G = \Delta H - T \Delta S \] where \(T\) is the temperature measured in Kelvin.
A negative \(\Delta G\) suggests that a reaction is spontaneous, meaning it can proceed without any external energy input. In the given exercise, calculation shows \(\Delta G = -285.8489 \, \mathrm{kJ \ mol}^{-1}\), meaning the reaction is spontaneous.

• If \(\Delta G\) is zero, the system is at equilibrium, and no net change occurs.
• A positive \(\Delta G\) indicates a non-spontaneous reaction, requiring energy input to proceed.

Overall, thermodynamics in chemistry helps chemists understand and predict the direction and extent of chemical reactions.