Question

Use data from Appendix \(\mathrm{C}\) to calculate the equilibrium constant, \(K\), at \(298 \mathrm{~K}\) for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Short answer

For the given reactions, the equilibrium constants K at 298 K are calculated as follows: (a) For the reaction H2(g) + I2(g) ⇌ 2 HI(g), K ≈ 7.04 × 10^10. (b) For the reaction C2H5OH(g) ⇌ C2H4(g) + H2O(g), first calculate ΔH°, ΔS°, and ΔG° using the data from Appendix C and then use K = e^{-ΔG°/ RT} to find K. (c) For the reaction 3 C2H2(g) ⇌ C6H6(g), first calculate ΔH°, ΔS°, and ΔG° using the data from Appendix C and then use K = e^{-ΔG°/ RT} to find K.

Step-by-step solution

(a) Calculating K for the reaction H2(g) + I2(g) ⇌ 2 HI(g)

Step 1: Calculate ΔH° using the enthalpy of formation of each substance involved in the reaction From Appendix C, we can find the standard enthalpy of formation for each substance: ΔH°(H2) = 0 kJ/mol (since it is an element in its standard state) ΔH°(I2) = 0 kJ/mol (since it is an element in its standard state) ΔH°(HI) = -26.48 kJ/mol Now we can calculate the ΔH° for the reaction: ΔH° = [2 × ΔH°(HI)] - [ΔH°(H2) + ΔH°(I2)] ΔH° = [2 × (-26.48 kJ/mol)] - [0] = -52.96 kJ/mol Step 2: Calculate ΔS° using the standard entropy of each substance involved in the reaction From Appendix C, we can find the standard entropy for each substance: S°(H2) = 130.7 J/mol·K S°(I2) = 260.7 J/mol·K S°(HI) = 206.6 J/mol·K Now we can calculate the ΔS° for the reaction: ΔS° = [2 × S°(HI)] - [S°(H2) + S°(I2)] ΔS° = [2 × 206.6 J/mol·K] - [130.7 J/mol·K + 260.7 J/mol·K] = 21.8 J/mol·K Step 3: Calculate ΔG° using the equation ΔG° = ΔH° - TΔS° At T = 298 K: ΔG° = -52.96 kJ/mol - [(298 K) × (21.8 J/mol·K × (1 kJ/1000 J))] ΔG° = -52.96 kJ/mol - 6.50 kJ/mol = -59.46 kJ/mol Step 4: Calculate K using the equation K = e^{-ΔG° / RT} K = e^{-(−59.46 kJ/mol) / [(8.314 × 10^{-3} kJ/mol·K) × (298 K)]} K = e^{24.97} K ≈ 7.04 × 10^10 Thus, for the reaction H2(g) + I2(g) ⇌ 2 HI(g), the equilibrium constant K at 298 K is approximately 7.04 × 10^10. Repeat the same steps for the other two reactions:

(b) Calculating K for the reaction C2H5OH(g) ⇌ C2H4(g) + H2O(g)

Step 1: Calculate ΔH° using the enthalpy of formation of each substance Step 2: Calculate ΔS° using the standard entropy of each substance Step 3: Calculate ΔG° using the equation ΔG° = ΔH° - TΔS° Step 4: Calculate K using the equation K = e^{-ΔG°/ RT} Note: Use the data from Appendix C for the substances involved in the reaction.

(c) Calculating K for the reaction 3 C2H2(g) ⇌ C6H6(g)

Step 1: Calculate ΔH° using the enthalpy of formation of each substance Step 2: Calculate ΔS° using the standard entropy of each substance Step 3: Calculate ΔG° using the equation ΔG° = ΔH° - TΔS° Step 4: Calculate K using the equation K = e^{-ΔG°/ RT} Note: Use the data from Appendix C for the substances involved in the reaction.

Key concepts

Standard Enthalpy of Formation

The standard enthalpy of formation, denoted as \( \Delta H^\circ_f \), is a crucial concept in chemistry, particularly when calculating the equilibrium constant. It represents the heat change when one mole of a compound is formed from its elements in their standard states, at 1 atmosphere of pressure and 298 K (25 °C).
Understanding \( \Delta H^\circ_f \) helps us determine whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).

• For example, elements like \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \) have a standard enthalpy of formation of 0 \( \mathrm{kJ/mol} \) because they are already in their standard states.
• In the reaction \( \mathrm{H}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \), the \( \mathrm{HI} \) compound, formed from these elements, has a \( \Delta H^\circ_f \) of -26.48 \( \mathrm{kJ/mol} \).

By calculating the sum of the products' enthalpies minus that of the reactants, we find the total enthalpy change \( \Delta H^\circ \) for the reaction. This gives insight into the energy changes accompanying the reaction.

Standard Entropy

Standard entropy is denoted by \( S^\circ \) and measures the randomness or disorder within a system.
The unit for standard entropy is \( \mathrm{J/mol\cdot K} \). Every substance has a unique standard entropy value, indicating the level of disorder compared to absolute zero, where entropy is zero.

• In the reaction of hydrogen and iodine to form hydrogen iodide ( \( \mathrm{H}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \)), the standard entropies \( S^\circ \) are \( 130.7 \mathrm{J/mol\cdot K} \), \( 260.7 \mathrm{J/mol\cdot K} \), and \( 206.6 \mathrm{J/mol\cdot K} \) for \( \mathrm{H}_2 \), \( \mathrm{I}_2 \), and \( \mathrm{HI} \) respectively.

By evaluating the difference in \( S^\circ \) between reactants and products, \( \Delta S^\circ \) can be calculated.
This value becomes important in determining the spontaneity and feasibility of a reaction when combined with \( \Delta H^\circ \) in Gibbs free energy calculations.

Gibbs Free Energy

Gibbs free energy, denoted as \( \Delta G^\circ \), integrates both enthalpy and entropy to predict reaction spontaneity.
It is described by the equation:\[\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ\]where \( T \) is the temperature in Kelvin.

• \( \Delta G^\circ < 0 \): Reaction is spontaneous.
• \( \Delta G^\circ = 0 \): Reaction is at equilibrium.
• \( \Delta G^\circ > 0 \): Reaction is non-spontaneous under standard conditions.

In our given example of forming \( \mathrm{HI} \) from \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \), the calculated \( \Delta G^\circ \) influences the magnitude of equilibrium constant \( K \).
A more negative \( \Delta G^\circ \) value indicates a larger \( K \), suggesting a greater extent to which the reaction proceeds in the forward direction.

Temperature Dependency

Temperature plays a significant role in chemical reactions, affecting both equilibrium and reaction rate.
Most importantly, the sign and magnitude of \( \Delta G^\circ \) change with temperature, thus altering the equilibrium constant \( K \) derived from:\[K = e^{-\Delta G^\circ / RT}\]where \( R \) is the universal gas constant.
As temperature increases or decreases, the entropy and enthalpy contributions can change the favorability of a reaction.

• If a reaction is endothermic (positive \( \Delta H^\circ \)), raising the temperature might increase \( K \).
• Conversely, for exothermic reactions, \( K \) might decrease as temperature rises.

Thus, it’s essential to know the temperature when predicting both the extent of reaction and its spontaneity.

Reaction Quotient

The reaction quotient, \( Q \), is similar to the equilibrium constant, but is calculated at any point during a reaction, not just at equilibrium.
It helps gauge how far a reaction has progressed and the direction in which it will proceed to reach equilibrium:

• \( Q = K \): The system is at equilibrium.
• \( Q < K \): The reaction will proceed forward to reach equilibrium.
• \( Q > K \): The reaction will proceed in reverse to reach equilibrium.

By comparing \( Q \) with \( K \), chemists can determine if adjustments are needed, such as changing concentrations or pressures to achieve equilibrium.
It's a vital tool in understanding real-time reaction dynamics, beyond the snapshot view provided by equilibrium constants alone.