Question

Using data from Appendix \(\mathrm{C}\) , calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at 298 \(\mathrm{K}\) under standard conditions. $$ \begin{array}{l}{\text { (a) } 2 \mathrm{Ag}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AgCl}(s)} \\ {\text { (b) } \mathrm{P}_{4} \mathrm{O}_{10}(s)+16 \mathrm{H}_{2}(g) \longrightarrow 4 \mathrm{PH}_{3}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (c) } \mathrm{CH}_{4}(g)+4 \mathrm{F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)} \\ {\text { (d) } 2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)}\end{array} $$

Short answer

For each reaction, we calculate \(\Delta G^{\circ}\) using the formula \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]. (a) \(\Delta G^{\circ}\)=-109.8 kJ/mol; spontaneous at 298 K (b) \(\Delta G^{\circ}\)=91.0 kJ/mol; nonspontaneous at 298 K (c) \(\Delta G^{\circ}\)=-216.2 kJ/mol; spontaneous at 298 K (d) \(\Delta G^{\circ}\)=46.9 kJ/mol; nonspontaneous at 298 K

Step-by-step solution

Find \(\Delta H^{\circ}\) for the reaction

Use the values for standard enthalpy of formation (\(\Delta H_f^{\circ}\)) for each compound from Appendix C. Subtract the sum of \(\Delta H_f^{\circ}\) of the reactants from the sum of \(\Delta H_f^{\circ}\) of the products.

Find \(\Delta S^{\circ}\) for the reaction

Use the values for standard entropy (\(S^{\circ}\)) for each compound from Appendix C. Subtract the sum of \(S^{\circ}\) of the reactants from the sum of \(S^{\circ}\) of the products.

Calculate \(\Delta G^{\circ}\) for the reaction

Use the formula \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] to calculate \(\Delta G^{\circ}\) for the reaction at 298 K.

Determine the spontaneity of the reaction

If \(\Delta G^{\circ}\) is negative, the reaction is spontaneous at 298 K under standard conditions. If \(\Delta G^{\circ}\) is positive, the reaction is nonspontaneous at that temperature. Repeat these steps for each reaction (a), (b), (c), and (d).

Key concepts

Standard Enthalpy of Formation

The Standard Enthalpy of Formation, denoted as \(\Delta H_f^{\circ}\), is a fundamental concept in thermodynamics. It represents the change in enthalpy when one mole of a compound is formed from its elements in their standard states. This measurement is crucial because it provides a baseline for predicting the heat absorbed or released during chemical reactions.

Key points to remember about \(\Delta H_f^{\circ}\):

• The values are usually expressed in kilojoules per mole (kJ/mol).
• The standard state of a substance is typically its most stable physical form at 1 atmosphere pressure and 298 K.
• Elements in their standard states have a \(\Delta H_f^{\circ}\) of zero.

In practical terms, calculating the standard enthalpy change (\(\Delta H^{\circ}\)) for a reaction involves using these values to determine how much energy is absorbed or released. This is found by subtracting the total \(\Delta H_f^{\circ}\) of reactants from that of the products. This step, often the first when determining Gibbs Free Energy, helps gauge the heat change in a given chemical reaction.

Standard Entropy

Standard Entropy, symbolized as \(S^{\circ}\), measures the amount of disorder or randomness in a system at standard conditions. It plays a vital role in understanding the spontaneity of reactions when exploring thermodynamic properties.

Important aspects of \(S^{\circ}\) include:

• It is typically expressed in joules per kelvin per mole (J/K·mol).
• Unlike enthalpy, even elements in their standard states have non-zero entropy values because a perfectly ordered state of absolute zero entropy only exists at absolute zero temperature, according to the third law of thermodynamics.
• Standard entropy values are used to calculate the entropy change (\(\Delta S^{\circ}\)) for a reaction, which reflects how the disorder or randomness in a system changes.

To find \(\Delta S^{\circ}\) for a reaction, subtract the sum of the standard entropies of the reactants from that of the products. This measurement helps predict whether the disorder increases or decreases as the reaction proceeds, providing insight into the system's spontaneity.

Spontaneity of Reactions

Understanding the spontaneity of reactions is crucial to predicting whether a chemical reaction can occur without external input. This determination is largely based on calculating the Gibbs Free Energy change, \(\Delta G^{\circ}\), a key thermodynamic quantity.

Here's how \(\Delta G^{\circ}\) relates to spontaneity:

• A negative \(\Delta G^{\circ}\) indicates that a reaction is spontaneous under standard conditions. This means the reaction will proceed without needing additional energy.
• A positive \(\Delta G^{\circ}\) suggests that a reaction is non-spontaneous, requiring energy input to proceed.
• Using the formula \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), you can integrate temperature effects, as shown by 298 K in standard conditions, to understand how both enthalpy and entropy influence spontaneity.

In essence, for a reaction to be spontaneous, the drive from increased disorder (entropy) must overcome the energy barrier (enthalpy) unless unfavorable conditions, such as significantly low temperature, impact the result. Calculating \(\Delta G^{\circ}\) allows us to consider these factors holistically, guiding predictions about reaction behavior.