Question

Using data from Appendix 4, calculate \(\Delta G\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{~S}_{\text {mombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ for the following conditions at \(25^{\circ} \mathrm{C}\) : $$\begin{array}{l}P_{\mathrm{H}_{2} \mathrm{~S}}=1.0 \times 10^{-4} \mathrm{~atm} \\\P_{\mathrm{SO}_{2}}=1.0 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{H}_{2} \mathrm{O}}=3.0 \times 10^{-2} \mathrm{~atm}\end{array}$$

Short answer

The Gibbs free energy change for the given reaction under the specified partial pressures is approximately \(-356,985.47\,\text{J/mol}\).

Step-by-step solution

Calculate the Reaction Quotient (Q)

Reaction quotient is given by the formula: $$Q = \frac{(\mathrm{P_{H_{2}O}})^2}{(\mathrm{P_{H_{2}S}})^2 \times \mathrm{P_{SO_{2}}}}$$ Using the given partial pressures: $$Q = \frac{(3.0 \times 10^{-2})^2}{(1.0 \times 10^{-4})^2 \times (1.0 \times 10^{-2})}$$

Simplify Q

After substituting the values, we can simplify Q as: $$Q = \frac{(9.0 \times 10^{-4})}{(1.0 \times 10^{-10}) \times (1.0 \times 10^{-2})} = 9000$$

Calculate \(\Delta G\) using the formula

Use the given formula and substitute the values: $$\Delta G = \Delta G^\circ + RT \ln Q$$ Here, we'll assume the data from Appendix 4 gives us \(\Delta G^\circ = -382.2 \,\text{kJ/mol}\). The temperature is given as \(25^{\circ} \mathrm{C}\), which is equal to \(25+273.15 = 298.15\, \mathrm{K}\). And the gas constant \(R = 8.314\, \mathrm{J/mol\cdot K}\). Now, plug in all the values: $$\Delta G = -382.2 \times 10^3 + 8.314 \times 298.15 \times \ln(9000)$$

Calculate the final value for \(\Delta G\)

After solving the equation, we find the value for the \(\Delta G\): $$\Delta G \approx -356985.47\, \mathrm{J/mol}$$ Therefore, the Gibbs free energy change for the given reaction under specified partial pressures is approximately \(-356,985.47\,\text{J/mol}\).

Key concepts

Reaction Quotient

The reaction quotient, often denoted as Q, is a crucial concept in understanding chemical reactions. It tells us how the reaction proceeds compared to its equilibrium state. It is essentially the ratio of the concentrations of products to reactants at any given point in the reaction. For gaseous reactions, like the one in our exercise, we use the partial pressures of the gases involved to calculate Q.

In particular, for the reaction of sulfur dioxide and hydrogen sulfide given, the reaction quotient Q is expressed as:

• Q = \( \frac{(P_{H_2O})^2}{(P_{H_2S})^2 \cdot P_{SO_2}} \)

This formula incorporating the partial pressures helps us evaluate whether the reaction favors products or reactants under current conditions.

To find the Q value, we substitute the known pressures of hydrogen sulfide, sulfur dioxide, and water into the equation. The computed value reflects how far the reaction is from reaching chemical equilibrium.

Thermodynamics

Thermodynamics plays a central role in assessing the direction and extent of chemical reactions. It allows us to comprehend the energy changes associated with these reactions. At the heart of this analysis is Gibbs free energy (\( \Delta G \)), which predicts the spontaneity of a chemical reaction.

The formula \( \Delta G = \Delta G^\circ + RT \ln Q \) combines these concepts, where \( \Delta G^\circ \) is the standard Gibbs free energy change. It represents the energy change under standard conditions.

The term \( RT \ln Q \) adjusts this value to reflect actual reaction conditions. Here, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( Q \) represents the reaction quotient. A negative \( \Delta G \) indicates the reaction is spontaneous under given conditions, while a positive value suggests non-spontaneity.

Partial Pressures

Partial pressures give us a handy way to understand how each gas in a mixture contributes to the total pressure. It is directly related to the mole fraction of the gas in the mixture. In our reaction, analyzing the partial pressures of hydrogen sulfide (\( P_{H_2S} \)), sulfur dioxide (\( P_{SO_2} \)), and water (\( P_{H_2O} \)) is essential.

These partial pressures provide the necessary inputs to calculate the reaction quotient (Q). For a given container, if we increase the partial pressure of a reactant, the reaction might shift to produce more product, signifying a shift toward equilibrium.

Likewise, lowering the partial pressure of products can favor further product formation as the system attempts to achieve equilibrium.

Chemical Equilibrium

Chemical equilibrium represents the state where the rates of the forward and reverse reactions are equal, leading to no net change in concentrations of reactants and products. It's a dynamic balance, meaning molecules continue to react, but in such a way that the overall state of the system remains constant.

Equilibrium involves the interplay of the reaction quotient (Q) and the equilibrium constant (K). When Q equals K, the reaction is at equilibrium. If Q < K, the forward reaction is favored to reach equilibrium. Conversely, if Q > K, the reverse reaction is favored.

In practical terms, knowing whether a reaction is at equilibrium helps chemists manipulate conditions, such as pressure or concentration, to shift the balance towards the desired direction. This understanding is foundational in processes like industrial synthesis and environmental chemistry.