Problem 127 - Chemistry The Molecular Science

Question

The standard equilibrium constant is \(2.1 \times 10^{9}\) for this reaction at \(25^{\circ} \mathrm{C}\) $$ \mathrm{Zn}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(\mathrm{aq}) $$ (a) Calculate \(\Delta_{r} G^{\circ}\) at this temperature. (b) If standard-state concentrations of the reactants and products are combined, in which direction will the reaction proceed? (c) Calculate \(\Delta_{\mathrm{r}} G\) when \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\right]=0.010 \mathrm{M},\left[\mathrm{Zn}^{2+}\right]=\) $$ 0.0010 \mathrm{M}, \text { and }\left[\mathrm{NH}_{3}\right]=3.5 \times 10^{-4} \mathrm{M} $$

Step-by-Step Solution

Answer

(a) \(-49.5\text{ kJ/mol}\); (b) Reaction towards right; (c) \(40.78\text{ kJ/mol}\).

Step 1: Calculate \( \Delta_{r}G^{\circ} \)

\( \Delta_{r}G^{\circ} \) (the standard Gibbs free energy change) is related to the equilibrium constant \( K \) by the equation: \[ \Delta_{r}G^{\circ} = -RT\ln K \]where \( R \) is the gas constant \( 8.314 \text{ J/mol K} \), and \( T \) is the temperature in Kelvin (\( 25^{\circ}\text{C} = 298 \text{ K} \)). Plugging in the values:\[ \Delta_{r}G^{\circ} = -(8.314 \text{ J/mol K})(298 \text{ K})\ln (2.1 \times 10^{9}) \]Calculating the natural logarithm of the equilibrium constant and multiplying by \( -RT \):\[ \Delta_{r}G^{\circ} = -49482 \text{ J/mol} \] or \( -49.5 \text{ kJ/mol} \).

Step 2: Determine Reaction Direction at Standard-State

If the initial concentrations of the reactants and products are at standard state (1 M), the reaction quotient \( Q = 1 \). Since \( K > Q \), the reaction will shift to the right, towards products to reach equilibrium.

Step 3: Calculate \( \Delta_{r}G \) for Given Conditions

To calculate \( \Delta_{r}G \) when the concentrations of species are not at standard state, use:\[ \Delta_{r}G = \Delta_{r}G^{\circ} + RT\ln Q \]where \( Q \) is the reaction quotient, calculated as follows:\[ Q = \frac{[\text{Zn}(\text{NH}_3)_4^{2+}]}{[\text{Zn}^{2+}][\text{NH}_3]^4} = \frac{0.010}{0.0010 \times (3.5 \times 10^{-4})^4} \]Calculating \( Q \):\[ Q = \frac{0.010}{0.0010 \times (1.5 \times 10^{-13})} \approx 6.6667 \times 10^{10} \]Now calculate \( \Delta_{r}G \):\[ \Delta_{r}G = -49482 + (8.314)(298)\ln(6.6667 \times 10^{10}) \]Calculate \( \Delta_{r}G \):\[ \Delta_{r}G \approx 40777 \text{ J/mol} \text{ or } 40.78 \text{ kJ/mol} \].

Step 4: Short Answer

(a) \(-49.5\text{ kJ/mol}\); (b) Reaction proceeds towards the right; (c) \(40.78\text{ kJ/mol}\).

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, often denoted as \( \Delta G \), is a crucial thermodynamic quantity used to predict the spontaneity of chemical reactions. It is a measure of the maximum reversible work that may be performed by a thermodynamic system at constant temperature and pressure. When calculating the standard Gibbs Free Energy change, \( \Delta_r G^{\circ} \), one typically relies on the relationship: \[ \Delta_r G^{\circ} = -RT\ln K \] where \( R \) is the universal gas constant \( (8.314 \text{ J/mol K}) \), \( T \) is the temperature in Kelvin, and \( K \) is the equilibrium constant of the reaction.

• If \( \Delta G < 0 \), the reaction is spontaneous under the given conditions.
• If \( \Delta G = 0 \), the system is at equilibrium.
• If \( \Delta G > 0 \), the reaction is non-spontaneous as written (but will proceed spontaneously in the reverse direction).

In essence, Gibbs Free Energy tells us about the feasibility and directionality of chemical processes, making it an indispensable tool for chemists when studying reactions at chemical equilibrium.

Reaction Quotient

The Reaction Quotient, \( Q \), serves as a snapshot of a reaction’s progress towards equilibrium. It is calculated using the same formula as the equilibrium constant \( K \), but at any point in time, not just equilibrium. For a typical reaction: \[ aA + bB \rightleftharpoons cC + dD \] the reaction quotient is expressed as:\[ Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]The value of \( Q \) compared to \( K \) helps to determine the direction in which the reaction will proceed to reach equilibrium:

• If \( Q < K \), the reaction will proceed forward, creating more products until equilibrium is reached.
• If \( Q = K \), the reaction is at equilibrium, meaning there is no net change in concentrations over time.
• If \( Q > K \), the reaction will proceed in reverse, forming more reactants until equilibrium is established.

In practical terms, \( Q \) helps chemists decide whether to expect more of the reactants or products in a given reaction setup.

Chemical Equilibrium

Chemical Equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, leading to no overall change in the concentrations of reactants and products. At this point, the reaction has reached a balance, characterized by an equilibrium constant \( K \).The dynamic nature of chemical equilibrium means that reactions continue to occur, but because they occur at the same rate in both directions, the concentrations remain constant. This concept is pivotal in understanding how reactions behave under various conditions. Key characteristics of chemical equilibrium include:

• The reaction does not stop; both forward and reverse reactions proceed at all times.
• The concentrations of reactants and products stay constant but not necessarily equal.
• External changes, such as temperature and pressure, can shift the equilibrium position, described by Le Chatelier's Principle.

Understanding chemical equilibrium is vital for predicting how a reaction can be driven in a desired direction by altering conditions, making it a fundamental concept in chemistry.