Problem 125 - Chemistry: An Atoms-Focused Approach

Question

Carbon disulfide is a foul-smelling solvent that dissolves sulfur and other nonpolar substances. It can be made by heating sulfur in an atmosphere of methane: $$4 \mathrm{CH}_{4}(g)+\mathrm{S}_{\mathrm{g}}(s) \longrightarrow 4 \mathrm{CS}_{2}(g)+8 \mathrm{H}_{2}(\mathrm{g})$$ Starting with the appropriate data in Appendix \(4,\) calculate the values of \(K_{\mathrm{p}}\) for the reaction at \(25^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

Question: Calculate the values of Kp for the given chemical reaction at 25°C and 500°C, using the standard Gibbs free energy of formation (∆Gf°) values and the relationship between ΔG° and Kp. Provide your answer in two decimal places. Answer: Kp at 25°C is ______, and Kp at 500°C is ______.

Step 1: Obtain standard Gibbs free energy of formation values from Appendix 4

You can find the standard Gibbs free energy of formation for each of the compounds involved in the reaction in Appendix 4. Write down the values of ∆Gf° for CH4(g), S8(s), CS2(g), and H2(g).

Step 2: Write the balanced chemical equation for the reaction

The balanced chemical equation is given in the exercise: $$4 \mathrm{CH}_{4}(g)+\mathrm{S}_{\mathrm{g}}(s) \longrightarrow 4 \mathrm{CS}_{2}(g)+8 \mathrm{H}_{2}(\mathrm{g})$$

Step 3: Calculate the standard Gibbs free energy change (ΔG°) for the reaction

Using the balanced chemical equation and the standard Gibbs free energy of formation (∆Gf°) values, we can calculate the standard Gibbs free energy change for the reaction as follows: $$\Delta G^\circ_\mathrm{rxn} = [\Delta G_f^\circ(\mathrm{CS_2}) \cdot 4 + \Delta G_f^\circ(\mathrm{H_2}) \cdot 8] - [\Delta G_f^\circ(\mathrm{CH_4}) \cdot 4 + \Delta G_f^\circ(\mathrm{S_8})]$$ Calculate the value of ΔG° for the reaction, using the values of ∆Gf° obtained in Step 1.

Step 4: Calculate Kp at 25°C using the relationship between ΔG° and Kp

We can find the equilibrium constant Kp for the reaction at 25°C using the relationship between ΔG° and Kp: $$\Delta G^\circ_\mathrm{rxn} = -RT \ln K_\mathrm{p}$$ Where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin (298.15 K for 25°C). Solve for Kp: $$K_\mathrm{p} = e^{-\frac{\Delta G^\circ_\mathrm{rxn}}{RT}}$$ Calculate Kp at 25°C using the value of ΔG° found in Step 3.

Step 5: Repeat Step 4 to find Kp at 500°C

To determine the equilibrium constant (Kp) at 500°C, we repeat Step 4 using the temperature 500°C, which is equal to 773.15 K: $$K_\mathrm{p} = e^{-\frac{\Delta G^\circ_\mathrm{rxn}}{RT}}$$ Calculate Kp at 500°C using the value of ΔG° found in Step 3 and the temperature value of 773.15 K.

Key Concepts

Gibbs Free Energy

Gibbs free energy, denoted as \( \Delta G \), is a fundamental concept in chemical thermodynamics that predicts the favorability of a chemical reaction. It combines enthalpy, entropy, and temperature to give a more comprehensive picture of a system's energy dynamics.

• If \( \Delta G \) is negative, the reaction is spontaneous under constant temperature and pressure conditions.
• If it's positive, the reaction is non-spontaneous and requires energy input to proceed.

The standard Gibbs free energy change \( \Delta G^\circ \) refers to these calculations under standard conditions (1 atm pressure and 298 K temperature). It can be determined using the formula:\[\Delta G^\circ_{\text{rxn}} = \sum \Delta G_f^\circ \text{(products)} - \sum \Delta G_f^\circ \text{(reactants)}\]where \( \Delta G_f^\circ \) is the standard Gibbs free energy of formation for each substance. By knowing these values for the components of a reaction, we can compute \( \Delta G^\circ \), which helps us understand the feasibility of the reaction.

Equilibrium Constant (Kp)

The equilibrium constant, \( K_p \), is a crucial concept that represents the ratio of product pressures to reactant pressures at chemical equilibrium for reactions involving gases. Arising from the law of mass action, \( K_p \) is dependent on the reaction's balanced equation and is expressed as:\[K_p = \frac{(P_{\text{CS}_2})^4 (P_{\text{H}_2})^8}{(P_{\text{CH}_4})^4 (P_{\text{S}_8})}\]where \( P \) denotes the partial pressures of each gas.The link between Gibbs free energy and \( K_p \) is given by:\[\Delta G^\circ_{rxn} = -RT \ln K_p\]Here, \( R \) is the universal gas constant (8.314 J/(mol·K)), \( T \) is temperature in Kelvin, and \( K_p \) quantifies the extent of the reaction at equilibrium. Given \( \Delta G^\circ \), this equation allows us to solve for \( K_p \) and gain insights into the concentration changes at equilibrium. Notably, a high \( K_p \) suggests a reaction that heavily favors products, whereas a low \( K_p \) favors reactants.

Temperature Dependence of Reactions

Temperature plays a pivotal role in shifting the chemical equilibrium because most reactions have varied Gibbs free energy values at different temperatures. This change is explained by the Gibbs-Helmholtz equation:\[\Delta G = \Delta H - T\Delta S\]where \( \Delta H \) is enthalpy change, \( T \) is temperature, and \( \Delta S \) is entropy change. Increasing temperature generally influences \( \Delta G \) and, consequently, the \( K_p \) value.

• For exothermic reactions, raising the temperature typically decreases \( K_p \) because \( \Delta G \) becomes less negative.
• In contrast, for endothermic reactions, higher temperatures often increase \( K_p \) since \( \Delta G \) becomes more negative.

Through temperature adjustments, we can predict how the equilibrium position of a reaction will shift—a key aspect for industrial applications where controlling reaction conditions maximizes yield.