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$$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \mapsto 2 \mathrm{NOCl}(g) \Delta G^{\circ}=132.6 \mathrm{kJ} / \mathrm{mol}$$ For the equilibrium above, what would happen to the value of \(\Delta G^{\circ}\) if the concentration of \(\mathrm{N}_{2}\) were to increase and why? (A) It would increase because the reaction would become more themodynamically favored. (B) It would increase because the reaction would shift right and create more products. (C) It would decrease because there are more reactants present. (D) It would stay the same because the value of \(K_{e q}\) would not change.

Short Answer

Expert verified
Answer: (D) The value of \(ΔG°\) would remain the same because it doesn't change with variations in concentration of the reactants, only the equilibrium position shifts, but the equilibrium constant \(K\) and hence \(ΔG°\) remain the same.

Step by step solution

01

Understanding Gibbs Free Energy and Equilibrium

First note, that the Gibbs Free Energy relationship with the equilibrium constant is given by the formula \(ΔG° = -RTlnK\), where \(ΔG°\) is the standard Gibbs free energy change, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(K\) is the equilibrium constant. Examining this equation tells us that at any given temperature \(T\), and \(R\) being a constant, \(ΔG°\) can only change if \(K\) changes.
02

Reactant Concentration and Equilibrium

The equilibrium constant \(K\) is determined by the specific reaction and the temperature, it is not influenced by the concentrations of the reactants or the products. In this exercise, the concentration of nitrogen (\(N_{2}\)) is supposed to be increased. However, this increase doesn't lead to a shift in the equilibrium constant \(K\), it only leads to a shift in the position of the equilibrium according to Le Chatelier's principle, by moving the equilibrium to the right.
03

Relation with Gibbs Free Energy

Since we established that an increase in the concentration of nitrogen will not change the equilibrium constant \(K\), it means that it also doesn't change the standard Gibbs free energy change \(ΔG°\). Any changes in reactant or product concentrations may shift the equilibrium position, but they do not alter the value of ∆G° for the reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
In chemical reactions, the equilibrium constant, denoted as \( K \), is vital in expressing the balance between reactants and products at equilibrium. It quantitatively describes the point where the rates of the forward and reverse reactions equalize. This constant is specific to each reaction and is only influenced by changes in temperature.

The equilibrium constant remains unchanged when you alter the concentration of reactants or products. This might seem unexpected, but it’s because \( K \) inherently depends on the ratio of product concentrations to reactant concentrations raised to certain powers as dictated by the balanced equation. Thus, even if the concentrations change, the system will adjust itself to maintain the same \( K \) value.
  • Remember, \( K \) changes only with temperature variations.
  • Alterations in pressure or concentration shift position but do not change \( K \).
Le Chatelier's Principle
Le Chatelier's Principle provides insight into how a system at equilibrium responds to disturbances or stresses. When a change such as concentration, pressure, or temperature is imposed on a reaction at equilibrium, the system will react in a way that minimizes the disturbance.

Imagine this as the system's way of 'compensating' for the change to re-establish equilibrium.
  • Increasing the concentration of a reactant leads to a shift towards the products, moving the equilibrium right.
  • Conversely, increasing the concentration of a product shifts equilibrium left, favoring more reactant formation.
This principle is pivotal in predicting how a reaction at equilibrium will respond to alterations, but it doesn't affect the equilibrium constant \( K \), only the concentrations of the involved substances.
Standard Gibbs Free Energy Change
The standard Gibbs free energy change, \( \Delta G^{\circ} \), is a key factor that determines the favorability and spontaneity of a chemical reaction under standard state conditions. It is directly linked to the equilibrium constant by the formula \( \Delta G^{\circ} = -RT\ln K \). This establishes that \( \Delta G^{\circ} \) changes only if \( K \) changes, which happens if there's a change in temperature.

This means that changes in concentrations of reactants or products at a set temperature will not alter \( \Delta G^{\circ} \), since \( K \) remains constant. In essence, \( \Delta G^{\circ} \) tells us the amount of energy change associated with transforming reactants into products under standard conditions without considering concentration variances.
  • If \( \Delta G^{\circ} \) is negative, the reaction is spontaneous under standard conditions.
  • A positive \( \Delta G^{\circ} \) indicates non-spontaneity.
Thermodynamic Favorability
Thermodynamic favorability is about whether a reaction can proceed on its own under a given set of conditions. A reaction is considered thermodynamically favorable when the driving forces, such as an entropic increase or enthalpy decrease, push it forward naturally.

This concept is closely linked with the Gibbs free energy change. When \( \Delta G \) is negative, the reaction releases free energy, indicating that it is favorable and might occur spontaneously. A positive \( \Delta G \) suggests that the reaction requires an input of energy to proceed.
  • In practice, for a reaction to be favorable, \( \Delta G \) must be less than zero.
  • Thermodynamic favorability doesn't always mean a reaction will occur rapidly; kinetics plays a role in speed.
Recognizing whether a reaction is favorable helps chemists understand potential pathways and reaction conditions.

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Most popular questions from this chapter

In a voltaic cell with a Cu(s) \(| \mathrm{Cu}^{2+}\) cathode and \(\mathrm{a} \mathrm{Pb}^{2+} | \mathrm{Pb}\) (s) anode, increasing the concentration of \(\mathrm{Pb}^{2+}\) causes the voltage to decrease. What is the reason for this? (A) The value for Q will increase, causing the cell to come closer to equilibrium. (B) The solution at the anode becomes more positively charged, leading to a reduced electron flow. (C) The reaction will shift to the right, causing a decrease in favor ability. (D) Cell potential will always decrease anytime the concentration of any aqueous species present increases.

$$4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ The above reaction will experience a rate increase by the addition of a catalyst such as platinum. Which of the following best explains why? (A) The catalyst causes the value for \(\Delta G\) to become more negative. (B) The catalyst increases the percentage of collisions that occur at the proper orientation in the reactant molecules. (C) The catalyst introduces a new reaction mechanism for the reaction. (D) The catalyst increases the activation energy for the reaction.

Use the following information to answer questions 12-15. When heated in a closed container in the presence of a catalyst, potassium chlorate decomposes into potassium chloride and oxygen gas via the following reaction: \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) If 12.25 g of potassium chlorate decomposes, how many grams of oxygen gas will be generated? (A) 1.60 g (B) 3.20 g (C) 4.80 g (D) 18.37 g

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. At the equivalence point, is the solution acidic, basic, or neutral? Why? (A) Acidic; the strong acid dissociates more than the weak base (B) Basic; the only ion present at equilibrium is the conjugate base (C) Basic; the higher concentration of the base is the determining factor (D) Neutral; equal moles of both acid and base are present

Which expression below should be used to calculate the mass of copper that can be plated out of a 1.0 \(\mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) , solution using a current of 0.75 A for 5.0 minutes? (A) \(\frac{(5.0)(60)(0.75)(63.55)}{(96500)(2)}\) (B) \(\frac{(5.0)(60)(63.55)(2)}{(0.75)(96500)}\) (C) \(\frac{(5.0)(60)(96500)(0.75)}{(63.55)(2)}\) (D) \(\frac{(5.0)(60)(96500)(63.55)}{(0.75)(2)}\)

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