Question

Consider the reaction: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2}(g)$$ where \(P_{\mathrm{NO}_{2}}=0.29\) atm and \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.6 \mathrm{atm} .\) For this reaction at these conditions, \(\Delta G=-1000 \mathrm{J}\) and \(\Delta G^{\circ}=6000 \mathrm{J}\) . Which of the following statements about this reaction is(are) true? a. The reverse reaction is spontaneous at these conditions. b. At equilibrium, \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}\) will be greater than 1.6 \(\mathrm{atm}\) . c. The value of K for this reaction is greater than 1. d. The maximum amount of work this reaction can produce at these conditions is –6000 J. e. The reaction is endothermic.

Short answer

The correct answer is: None of the given statements are true. The reaction is spontaneous (forward direction). The value of K at these conditions is 0.057, which is less than 1, and the reaction is exothermic. Therefore, all given statements are false.

Step-by-step solution

Determine the spontaneity of the reaction

The spontaneity of a reaction can be determined using the value of ΔG. If ΔG is negative, the reaction is spontaneous, and if ΔG is positive, the reaction is non-spontaneous. For this reaction, \(\Delta G=-1000 \mathrm{J}\). Since ΔG is negative, the reaction is spontaneous.

Calculate the reaction quotient Q

Let's calculate the reaction quotient (Q) using the given partial pressures of the reactants and products. $$Q = \frac{(P_{NO_2})^2}{P_{N_2O_4}}$$ Plug in the given values: $$Q = \frac{(0.29 \mathrm{atm})^2}{1.6 \mathrm{atm}}$$ $$Q = 0.0521$$

Find the value of K using ΔG and ΔG°

We can find the equilibrium constant K using the formula: $$\Delta G = \Delta G^{\circ} + RT\ln{Q} $$ where R is the gas constant (8.314 J/(mol∙K)) Since ΔG < 0, we can rewrite the formula: $$\Delta G^{\circ} - \Delta G = - RT\ln{Q} $$ And we can find K using ΔG° and the formula: $$\Delta G^{\circ} = - RT\ln{K} $$ Now plug in the values: $$6000\mathrm{J} = 1000\mathrm{J} + 8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot T\ln{Q} $$ $$5000\mathrm{J} = 8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot T\ln{Q} $$ Now, we can find the value of K using the relation: $$\ln{K} = \frac{\Delta G^{\circ}}{- RT} $$ Plug in the values: $$\ln{K} = \frac{6000\mathrm{J}}{-8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot T} $$ We need to find the temperature first to get the value of K. To find the temperature, we use equation $$5000\mathrm{J} = 8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot T\ln{0.0521} $$ $$ T = \frac{5000\mathrm{J}}{{-8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot \ln{0.0521}} = 236.82 \,\mathrm{K}$$ Now plug back the Temperature and find K: $$\ln{K} = \frac{6000\mathrm{J}}{-8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot 236.82 \,\mathrm{K}} $$ $$\ln{K} = -2.86$$ $$K = e^{-2.86} = 0.057$$

Analyze the given statements

a. The reverse reaction is spontaneous at these conditions. False. Since ΔG is negative, the forward reaction is spontaneous. b. At equilibrium, \(P_{N_{2}O_{4}}\) will be greater than 1.6 atm. False. Since K < 1, the value of \(P_{N_{2}O_{4}}\) will be less than 1.6 atm at equilibrium. c. The value of K for this reaction is greater than 1. False. We calculated the value of K to be 0.057, which is less than 1. d. The maximum amount of work this reaction can produce at these conditions is –6000 J. False. The maximum amount of work that can be produced is given by ΔG, which is -1000 J. e. The reaction is endothermic. False. Since ΔG° is positive and K < 1, the reaction is exothermic. Based on our analysis, all of the given statements are false.

Key concepts

Reaction Quotient

The reaction quotient (Q) is used to determine the direction a chemical reaction will proceed to reach equilibrium. It's essentially a snapshot of the current situation in the reaction, comparing the concentrations of products and reactants at any given time. To find Q, use the equation:
\[ Q = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2O_4}} \]Here, Q is calculated using the pressures of the gaseous chemicals involved. For the reaction \(\text{N}_2\text{O}_4(g) \rightarrow 2\text{NO}_2(g)\), the given values are \(P_{\text{NO}_2} = 0.29\text{ atm}\) and \(P_{\text{N}_2O_4} = 1.6\text{ atm}\). Plugging these in, Q is found to be 0.0521. Comparing Q to the equilibrium constant (K) helps to predict the shift needed to reach equilibrium:

• If Q < K, the reaction moves forward, producing more products.
• If Q > K, the reaction shifts backward, increasing reactant concentrations.
• If Q = K, the system is at equilibrium.

Equilibrium Constant

The equilibrium constant (K) gives insight into the position of equilibrium for a reaction at a particular temperature. It is calculated when a reaction reaches a state where the rate of the forward reaction equals the rate of the reverse one, and the concentrations of reactants and products remain constant over time.
For the reaction given, K is calculated using the Gibbs Free Energy equation \(\Delta G^\circ = - RT \ln K\). We determined the K value to be approximately 0.057. This low value indicates that at equilibrium, the reaction favors the reactants (\(\text{N}_2\text{O}_4\)) rather than the products (\(\text{NO}_2\)).This means that, under the given conditions, the reaction does not proceed very far towards products and remains more towards the reactants, highlighting its equilibrium state.

Thermodynamics

Thermodynamics in chemistry relates to the study of energy changes involved in chemical reactions. A key component is the Gibbs Free Energy (ΔG), which tells us about the spontaneity and potential work a reaction can perform.
Delta G \((\Delta G^\circ)\) incorporates both the system's enthalpy and entropy changes with the situation's temperature, helping to determine reaction feasibility.

• If ΔG < 0, the reaction is spontaneous; it can proceed without outside energy.
• If ΔG > 0, the reaction is non-spontaneous; it requires energy to proceed.
• If ΔG = 0, the system is at equilibrium.

In this exercise, ΔG is -1000 J, indicating a spontaneous forward reaction under these conditions, whereas ΔG^\circ is 6000 J, telling us about its standard state energetics. Understanding these parameters allows students to predict how a reaction will behave under varying conditions.

Reaction Spontaneity

Reaction spontaneity indicates whether a reaction will naturally proceed without external energy input. It's determined by the sign of Gibbs Free Energy (ΔG).
In this exercise:

• Given ΔG = -1000 J implies that the forward reaction (\(\text{N}_2\text{O}_4(g) \rightarrow 2 \text{NO}_2(g)\)) is spontaneous at the specified conditions.
• Conversely, if ΔG were positive, the reverse would be true, and the reaction would not occur without additional energy.
• Understanding reaction spontaneity helps predict reaction pathways and optimize conditions for desired product formation. ΔG is a vital tool in evaluating reaction feasibility and must consider not only the starting and ending states but the journey between these states as well.