Question

What is the approximate value of \(\log K_{\mathrm{p}}\) for the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(25^{\circ} \mathrm{C}\) The standard enthalpy of formation of \(\mathrm{NH}_{3}(\mathrm{~g})\) is \(-40.0 \mathrm{~kJ} / \mathrm{mol}\) and standard entropies of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{NH}_{3}(\mathrm{~g})\) are 191,130 and \(192 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), respectively. (a) \(0.04\) (b) \(7.05\) (c) \(8.6\) (d) \(3.73\)

Short answer

The approximate value of \(\log K_{\text{p}}\) is (d) 3.73.

Step-by-step solution

Calculate the standard reaction entropy, \(\Delta S^\circ_{\text{rxn}}\)

To find the standard reaction entropy, \(\Delta S^\circ_{\text{rxn}}\), subtract the sum of the standard molar entropies of the reactants from the sum of the standard molar entropies of the products. For the given reaction, the entropy change is calculated using the formula: \[\Delta S^\circ_{\text{rxn}} = (2 \times S^\circ_{\text{NH}_3}) - (S^\circ_{\text{N}_2} + 3 \times S^\circ_{\text{H}_2})\] where the standard molar entropies \((S^\circ)\) are given as 192 \(\text{J/K} \cdot \text{mol}\) for \(\text{NH}_3\), 191 \(\text{J/K} \cdot \text{mol}\) for \(\text{N}_2\), and 130 \(\text{J/K} \cdot \text{mol}\) for \(\text{H}_2\).

Calculate the standard reaction enthalpy, \(\Delta H^\circ_{\text{rxn}}\)

Calculate the standard reaction enthalpy, \(\Delta H^\circ_{\text{rxn}}\), by multiplying the enthalpy of formation of ammonia by 2 (since 2 moles of \(\text{NH}_3\) are produced) and subtracting the enthalpy of formation of reactants (which is 0 for \(\text{N}_2\) and \(\text{H}_2\) as they are in their standard states). The enthalpy change is: \[\Delta H^\circ_{\text{rxn}} = 2 \times (-40.0 \text{ kJ/mol}) - (0 + 0) = -80.0 \text{ kJ/mol}\]

Convert \(\Delta H^\circ_{\text{rxn}}\) to joules

The \(\Delta H^\circ_{\text{rxn}}\) found in Step 2 is in kilojoules. To convert it to joules, which is the unit of entropy, multiply by 1000: \[\Delta H^\circ_{\text{rxn}} = -80.0 \text{ kJ/mol} \times 1000 \text{ J/kJ} = -80,000 \text{ J/mol}\]

Calculate the equilibrium constant (\(K_{\text{p}}\)) at \(25^\circ C\)

Use the Gibbs free energy equation: \[\Delta G^\circ_{\text{rxn}} = \Delta H^\circ_{\text{rxn}} - T\Delta S^\circ_{\text{rxn}}\] where \(T\) is 298 K \(25^\circ C = 298 K\). The equilibrium constant \(K_{\text{p}}\) is found using the relation: \[\Delta G^\circ_{\text{rxn}} = -RT\ln K_{\text{p}}\] Rearrange for \(\text{K}_{\text{p}}\): \[K_{\text{p}} = e^{-\Delta G^\circ_{\text{rxn}} / (RT)}\] where \(R = 8.314 \text{ J/(K \cdot mol)}\).

Calculate \(\log K_{\text{p}}\)

The natural logarithm \(\ln K_{\text{p}}\) can be converted to a base-10 log using the conversion formula: \[\log K_{\text{p}} = \ln K_{\text{p}} / \ln(10)\] Calculate the value of \(\log K_{\text{p}}\) to determine the approximate value as given in options (a) to (d).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when a chemical reaction and its reverse reaction proceed at the same rate. As a result, the concentrations of reactants and products remain unchanged over time, creating a state of dynamic balance. This concept is crucial to understand because it dictates how chemicals react and at what concentrations the products and reactants exist when a reaction is left to proceed until it's no longer changing.

When calculating equilibrium constants, we are essentially quantifying this balance. For gas-phase reactions, we use the equilibrium constant in terms of pressure, denoted as \( K_{\mathrm{p}} \). It is related to the concentration-based constant, \( K_{\mathrm{c}} \), through the ideal gas law. In the exercise given, we're looking for the \( \log K_{\mathrm{p}} \) of the synthesis of ammonia, which can tell us much about the reaction's favorability under standard conditions.

Standard Enthalpy of Formation

The standard enthalpy of formation, \( \Delta H_{\mathrm{f}}^\circ \), is a measure of the energy change when one mole of a substance is formed from its elements in their standard states. The standard state of a substance is its pure form at 1 bar of pressure and a specified temperature, typically \( 25^\circ C \) or \( 298 K \).

When a chemical reaction occurs, the standard enthalpy of formation for the products and the reactants come into play to determine the overall energy change, or standard reaction enthalpy, \( \Delta H_{\mathrm{rxn}}^\circ \). This value is found by summing the enthalpies of formation for the products and subtracting the sum of the enthalpies of formation of the reactants. In the case of the synthesis of ammonia from nitrogen and hydrogen gases, the \( \Delta H_{\mathrm{f}}^\circ \) for the reactants is zero because they are in their elemental, stable forms. The enthalpy of formation is only provided for ammonia, the product, allowing us to calculate the total enthalpy change for the reaction.

Standard Reaction Entropy

Entropy, symbolized by \( S \), is a measure of the disorder or randomness of a system. The standard reaction entropy, \( \Delta S_{\mathrm{rxn}}^\circ \), represents the change in entropy for a reaction carried out at standard conditions (1 bar and \( 298 K \)).

To determine \( \Delta S_{\mathrm{rxn}}^\circ \), we subtract the sum of the standard molar entropies of the reactants from the sum of the standard molar entropies of the products, as illustrated in the solution steps. The entropy values provided for the reactants (\( \mathrm{N}_{2}(\mathrm{~g}) \) and \( \mathrm{H}_{2}(\mathrm{~g}) \)) and the product (\( \mathrm{NH}_{3}(\mathrm{~g}) \)) are essential to determining how the disorder of the system changes from reactants to products. An increase in entropy usually indicates more disorder, while a decrease implies a more ordered system.

Gibbs Free Energy

Gibbs free energy (\( \Delta G \)) is a thermodynamic quantity that serves as a criterion for spontaneity in a closed system at constant temperature and pressure. It combines the system's enthalpy and entropy into one value to predict whether a process will occur spontaneously. The relationship is given by the equation:
\[ \Delta G = \Delta H - T\Delta S \]
Where \( \Delta G \) is the change in Gibbs free energy, \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.

In a chemical reaction, a negative \( \Delta G \) indicates that the reaction can occur spontaneously under standard conditions. In the exercise, we use the standard reaction enthalpy and standard reaction entropy to determine the Gibbs free energy change for the reaction. This value then helps us find the equilibrium constant \( K_{\mathrm{p}} \) at a given temperature, which can be transformed into the logarithmic form, \( \log K_{\mathrm{p}} \), providing a more intuitive sense of the equilibrium constant's magnitude.