Problem 68 - Chemistry An Atoms First Approach

Question

Consider the following reaction: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of ±1 in all quantities): a. \(T=298 \mathrm{K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{atm}\) b. \(T=298 \mathrm{K}, P_{\mathrm{N}_{2}}=200 \mathrm{atm}, P_{\mathrm{H}_{2}}=600 \mathrm{atm}\) \(P_{\mathrm{NH}_{3}}=200 \mathrm{atm}\)

Step-by-Step Solution

Answer

For the given reaction and conditions: a) ΔG = ΔG° + (8.314 J/mol·K)(298 K) ln(\(\frac{50^2}{200 \cdot 200^3}\)) Based on the reaction quotient (Q), we can determine that the reaction will be spontaneous or non-spontaneous in the forward direction for case 'a'. b) ΔG = ΔG° + (8.314 J/mol·K)(298 K) ln(\(\frac{200^2}{200 \cdot 600^3}\)) Similarly, we can determine the spontaneity of the reaction for case 'b' based on the reaction quotient (Q). Exact values for ΔG in both scenarios cannot be calculated due to insufficient information on standard Gibbs free energy of formation (ΔG°) and equilibrium constant (K).

Step 1: Write down the Van't Hoff equation

The equation for calculating Gibbs free energy change (ΔG) in a reaction is: \[\Delta G = \Delta G^\circ + RT \ln Q\] where ΔG° is standard Gibbs free energy of formation, R is the universal gas constant, T is the temperature, and Q is the reaction quotient.

Step 2: Calculate the equilibrium constant (K)

The equilibrium constant (K) is related to the standard Gibbs free energy of formation (ΔG°) by the following equation: \[K = e^{-\frac{\Delta G^\circ}{RT}}\] However, in this problem, we aren't given values for ΔG°, thus we cannot determine K directly.

Step 3: Calculate the reaction quotient (Q)

The reaction quotient (Q) is calculated by using the partial pressures of the reactants and products. For this reaction: \[Q = \frac{P_{NH_3}^2}{P_{N_2} \cdot P_{H_2}^3}\] Let's calculate Q for both sets of conditions: a) T = 298 K, PN2 = PH2 = 200 atm, PNH3 = 50 atm \[Q_a = \frac{50^2}{200 \cdot 200^3}\] b) T = 298 K, PN2 = 200 atm, PH2 = 600 atm, PNH3 = 200 atm \[Q_b = \frac{200^2}{200 \cdot 600^3}\]

Step 4: Calculate ΔG for each set of conditions

Since we don't have exact values for K or ΔG°, we will at least identify whether the reaction will be spontaneous or not spontaneous in each scenario: ΔG = ΔG° + RT ln Q a) If Q < K, then ln(Q) < 0, and ΔG < ΔG° (reaction will be spontaneous in the forward direction) b) If Q > K, then ln(Q) > 0, and ΔG > ΔG° (reaction will be non-spontaneous in the forward direction) Plug in the calculated values of the Q into the ΔG equation: a) ΔG = ΔG° + (8.314 J/mol·K)(298 K) ln(\(\frac{50^2)}{200 \cdot 200^3}\)) Using the condition for spontaneity, we can conclude if the reaction is spontaneous or non-spontaneous for case 'a' without knowing the exact values. b) ΔG = ΔG° + (8.314 J/mol·K)(298 K) ln(\(\frac{200^2}{200 \cdot 600^3}\)) Similarly, using the condition for spontaneity, we can conclude if the reaction is spontaneous or non-spontaneous for case 'b'. Due to the lack of essential information such as the standard Gibbs free energy of formation (ΔG°) and the equilibrium constant (K), we are unable to calculate the exact values for ΔG in both scenarios. However, by calculating the reaction quotient (Q) and comparing it with the equilibrium constant (K), we are still able to determine the spontaneity of the reaction.

Key Concepts

Reaction Quotient

In chemistry, the reaction quotient, denoted as \( Q \), helps us understand the progress and direction of a chemical reaction at any given point. It is a snapshot of the ratio of product concentrations to reactant concentrations at a specific time, under non-equilibrium conditions. This value can help us determine how far our system is from reaching equilibrium.
To calculate \( Q \) for a reaction, we use the partial pressures or concentrations of the gases involved in a balanced chemical equation. Take the example reaction \( \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \). The formula to find \( Q \) is:
\[Q = \frac{P_{\text{NH}_3}^2}{P_{\text{N}_2} \cdot P_{\text{H}_2}^3} \]
where \( P \) denotes the partial pressure of each gas.
In essence, \( Q \) gives us a value to compare with the equilibrium constant to assess the reaction's directionality.

Equilibrium Constant

The equilibrium constant, represented as \( K \), is a key concept in understanding the balance of chemical reactions. It is a fixed value for a specific reaction at a given temperature and provides insight into which direction a reaction predominantly lies when it reaches equilibrium.
For the reaction \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \), \( K \) is calculated using the same formula as \( Q \), but with the concentrations at equilibrium:
\[K = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]
In situations where \( Q eq K \), it indicates that the reaction is still progressing towards equilibrium, either favoring product formation (if \( Q < K \)) or reactant reformation (if \( Q > K \)).
In this exercise, without being given \( \Delta G^\circ \) or \( K \), the exact position of equilibrium is unclear, but we can still use \( Q \) to understand reaction tendencies.

Reaction Spontaneity

The concept of reaction spontaneity is centered around whether a chemical process will occur on its own without the input of external energy. This determination involves Gibbs free energy change, \( \Delta G \), which must be negative for a process to be spontaneous.
The relationship between \( \Delta G \), the reaction quotient \( Q \), and equilibrium constant \( K \) is described by the equation:
\[\Delta G = \Delta G^\circ + RT \ln Q\]
where \( R \) is the universal gas constant and \( T \) is temperature in Kelvin. When \( Q < K \), \( \ln(Q/K) < 0 \) making \( \Delta G < 0 \) and implying a spontaneous process. Conversely, if \( Q > K \), \( \Delta G > 0 \) and the reaction is not spontaneous.
This exercise highlights how \( Q \) values are used to infer the spontaneity of a reaction by comparison with \( K \), guiding us even when exact numerical values of \( \Delta G \) aren't available.