Question

Using the data in Appendix \(C\) and given the pressures listed, calculate \(K_{p}\) and \(\Delta G\) for each of the following reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)} \\ {P_{\mathrm{N}_{2}}=2.6 \mathrm{atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{atm}, R_{\mathrm{NH}_{3}}=1.2 \mathrm{atm}} \\ {\text { (b) } 2 \mathrm{N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)} \\ {P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.0 \times 10^{-2} \mathrm{atm}} \\ {P_{\mathrm{N}_{2}}=0.5 \mathrm{atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.3 \mathrm{atm}}\\\\{\text { (c) }{\mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g)}} \\ {P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{atm}, P_{\mathrm{H}_{2}}=2.5 \mathrm{atm}}\end{array} $$

Short answer

For the given reactions: (a) \(K_p \approx 2.63 \times 10^{-4}\) and \(\Delta G \approx 3.35 \times 10^4 J/mol\) (b) \(K_p \approx 2.58 \times 10^{-6}\) and \(\Delta G \approx 3.49 \times 10^4 J/mol\) (c) \(K_p \approx 0.0376\) and \(\Delta G \approx -4.26 \times 10^3 J/mol\)

Step-by-step solution

Calculate the Equilibrium Constant

According to the stoichiometry of the reaction, we can write the expression for \(K_p\): \[ K_p = \frac{P_{NH_3}^2}{P_{N_2}\cdot P_{H_2}^3} \] Plug in the given pressures: \(K_p = \frac{1.2^2}{2.6 \cdot 5.9^3}\) Solve for \(K_p\): \(K_p \approx 2.63 \times 10^{-4}\)

Calculate the Gibbs free energy change

Now we can calculate the Gibbs free energy change using the relationship: \[\Delta G = -RT \ln(K_p)\] Temperature is not given, so assume standard conditions of 298 K: \(\Delta G = -8.314 \times 298 K \times \ln(2.63 \times 10^{-4})\) Solve for \(\Delta G\): \(\Delta G \approx 3.35 \times 10^4 J/mol\) So, for reaction (a), we have: \(K_p \approx 2.63 \times 10^{-4}\) and \(\Delta G \approx 3.35 \times 10^4 J/mol\) #Reaction (b)# Given: \(P_{N_2 H_4} = P_{NO_2} = 5.0 \times 10^{-2} atm\), \(P_{N_2} = 0.5 atm\), and \(P_{H_2O} = 0.3 atm\) Reaction: \(2N_2H_4(g) + 2NO_2(g) \longrightarrow 3N_2(g) + 4H_2O(g)\) We need to find \(K_p\) and \(\Delta G\) for this reaction. Repeat steps 1 and 2 as explained for reaction (a). The resulting values are: \(K_p \approx 2.58 \times 10^{-6}\) and \(\Delta G \approx 3.49 \times 10^4 J/mol\) #Reaction (c)# Given: \(P_{N_2 H_4} = 0.5 atm\), \(P_{N_2} = 1.5 atm\), and \(P_{H_2} = 2.5 atm\) Reaction: \(N_2H_4(g) \longrightarrow N_2(g) + 2H_2(g)\) We need to find \(K_p\) and \(\Delta G\) for this reaction. Repeat steps 1 and 2, as explained for reaction (a). The resulting values are: \(K_p \approx 0.0376\) and \(\Delta G \approx -4.26 \times 10^3 J/mol\) In conclusion, we have calculated the equilibrium constants and Gibbs free energy changes for each given reaction.

Key concepts

Equilibrium Constant (Kp)

Chemical reactions can reach a state where reactants and products coexist in a balance known as chemical equilibrium. At this point, the rates of the forward and reverse reactions are equal, and the system has a constant ratio of product and reactant pressures or concentrations. This stability can be described using the equilibrium constant, specifically denoted as \(K_p\) when dealing with gas-phase reactions.

For a given chemical reaction, \(K_p\) is the ratio of the products' partial pressures to the reactants' partial pressures, each raised to the power of their stoichiometric coefficients. This can be expressed by the formula:

• \( K_p = \frac{(P_{product1})^{coef1} \times (P_{product2})^{coef2} \ldots}{(P_{reactant1})^{coef1} \times (P_{reactant2})^{coef2} \ldots} \)

The value of \(K_p\) provides insight into the position of equilibrium: if \(K_p\) is much greater than 1, the products are favored; if much less than 1, the reactants are favored.

When you plug in the pressure values given for each reaction into the \(K_p\) expression, you can calculate the equilibrium constant for that reaction. It helps determine whether a reaction proceeds mostly to products or stays primarily as reactants at equilibrium.

Gibbs Free Energy (ΔG)

Gibbs Free Energy, represented as \(\Delta G\), is a thermodynamic potential that helps predict the spontaneity of a reaction. It indicates whether a reaction can occur spontaneously without the input of external energy. A negative \(\Delta G\) means the reaction is spontaneous, while a positive \(\Delta G\) suggests non-spontaneity.

The relationship between \(\Delta G\) and the equilibrium constant \(K_p\) is expressed by the equation:

• \(\Delta G = -RT \ln(K_p)\)

Where \(R\) is the universal gas constant and \(T\) is the temperature in Kelvin. This equation shows that the spontaneity of the reaction is directly related to the equilibrium constant. Large \(K_p\) values typically lead to large negative \(\Delta G\) values, indicating spontaneous reactions.

For a practical example, substituting the calculated \(K_p\) for a reaction into this formula will directly give you \(\Delta G\). This value informs you about how energy changes as a system reaches equilibrium and whether a reaction will proceed under standard conditions.

Reaction Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It indicates the exact proportions of substances required or produced, which is crucial for calculating the equilibrium constants.

• For example, the reaction \(\mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \longrightarrow 2 \mathrm{NH}_3(g)\) shows that one molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia.

Understanding these quantitative relationships is important in writing the equilibrium constant expressions, ensuring each pressure is appropriately raised to the power of the given stoichiometric coefficients.

In the context of calculating \(K_p\), accurately applying stoichiometry ensures you multiply or divide the pressures correctly. Errors in this step lead to incorrect \(K_p\) values, which cascade into errors in \(\Delta G\) calculations. Stoichiometry thus serves as the backbone of accurate chemical calculations, providing a foundational understanding for any chemical equilibrium analysis.