Problem 103 - Chemical Principles: The Quest for Insight

Question

At \(1565 \mathrm{~K}\), the equilibrium constants for the reactions (1) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) and \((2)\) \(2 \mathrm{CO}_{2}(\mathrm{~g})=2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) are \(1.6 \times 10^{-11}\) and \(1.3 \times 10^{-10}\), respectively. (a) What is the equilibrium constant for the reaction (3) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})=\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\) at that temperature? (b) Show that the manner in which equilibrium constants are calculated is consistent with the manner in which the \(\Delta G_{1}^{\circ}\) values are calculated when combining two or more equarions by determining \(\Delta G_{e}{\underline{\phantom{xx}}}^{\circ}\) for \((1)\) and \((2)\) and using those values to calculare \(\Delta G,{ }^{\circ}\) and \(K_{3}\) for reaction (3).

Step-by-Step Solution

Answer

The equilibrium constant for reaction (3) at 1565 K is \( K_{3} = 8.125 \times 10^{10} \) and the calculation of equilibrium constants is consistent with the standard Gibbs free energy changes when combining reactions as \( \Delta G_{3}^{\circ} \) calculated from \( \Delta G_{1}^{\circ} \) and \( \Delta G_{2}^{\circ} \) gives the same \( K_{3} \) value.

Step 1: Deriving the Combined Reaction

To find the equilibrium constant for reaction (3), we can combine reaction (1) and reaction (2). Reverse reaction (1) and add it to reaction (2) to get reaction (3). This involves reversing the direction of reaction (1), which will invert its equilibrium constant, and then adding it to reaction (2).

Step 2: Calculate the Equilibrium Constant for Reaction (3)

The equilibrium constant for the reverse of reaction (1) is the reciprocal of the equilibrium constant for the forward reaction, so for the reversed reaction (1) it is \( K_{1}^{\prime} = 1 / (1.6 \times 10^{-11}) \). The equilibrium constant for the combination of reversed reaction (1) and reaction (2) is the product of their equilibrium constants, \( K_{3} = K_{1}^{\prime} \times K_{2} = (1 / (1.6 \times 10^{-11})) \times (1.3 \times 10^{-10}) \).

Step 3: Determining the Standard Gibbs Free Energy Changes

To show consistency, calculate the standard Gibbs free energy change for reactions (1) and (2) using \( \Delta G^{\circ} = -RT \ln(K) \) for each reaction at equilibrium. Then, use the equation \( \Delta G_{3}^{\circ} = \Delta G_{2}^{\circ} - \Delta G_{1}^{\circ} \) to find the standard Gibbs free energy change for reaction (3). Convert \( \Delta G_{3}^{\circ} \) back to an equilibrium constant using \( K_{3} = e^{-\Delta G_{3}^{\circ} / (RT)} \) to confirm the equilibrium constant obtained in step 2.

Key Concepts

Chemical Equilibrium

In chemistry, chemical equilibrium refers to the state in which the concentrations of reactants and products remain constant over time because the forward and reverse reactions occur at the same rate. It's important to realize that this doesn't mean the reactants and products are in equal concentrations; rather, they have reached a balance of exchange.

When we write a chemical equation, such as

\[2 \mathrm{H}_2\mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g})\]

the double arrows signify that the reaction can proceed in both the forward and reverse directions. At equilibrium, the rate of the forward reaction (water decomposing into hydrogen and oxygen gas) is equal to the rate of the reverse reaction (hydrogen and oxygen gas combining to form water).

To understand how reactions behave when conditions change, we study their equilibrium constants, which provide valuable information about the reaction's position at equilibrium and how it will adjust in response to external changes.

Gibbs Free Energy

The concept of Gibbs free energy (G) is pivotal in predicting the spontaneity of chemical reactions. It is a thermodynamic quantity that combines enthalpy, entropy, and temperature to predict the favored direction of a chemical process. The change in Gibbs free energy, denoted as \(\Delta G\), during a reaction provides insights into whether the reaction can occur spontaneously.

The equation

\[\Delta G=\Delta H - T\Delta S\]

explains that a process will be spontaneous if it releases heat (exothermic, \(\Delta H < 0\)) and/or increases the entropy of the system (\(\Delta S > 0\)), at constant temperature (T).

In the context of chemical equilibria, we specifically look at the standard Gibbs free energy change (\(\Delta G^{\circ}\)). It is related to the equilibrium constant (K) by the equation

\[\Delta G^{\circ} = -RT\ln(K)\]

where R is the universal gas constant and T is the temperature in kelvins. If \(\Delta G^{\circ} < 0\), the reaction will tend to move towards the products (forward reaction), whereas if \(\Delta G^{\circ} > 0\), the reaction favors the reactants (reverse reaction). At equilibrium, \(\Delta G^{\circ} = 0\), meaning the system is at its maximum stability, and there's no net change in the composition of the system.

Le Chatelier’s Principle

When a system at equilibrium is subjected to an external change, such as variations in concentration, pressure, or temperature, the equilibrium position shifts to counteract the change - this is known as Le Chatelier's principle.

For instance, adding more reactants to the system can cause the equilibrium to shift to the right, meaning more products will be formed to re-establish equilibrium. Conversely, increasing the temperature of an exothermic reaction will typically shift the equilibrium to the left, as the system tries to absorb the added heat by favoring the reverse reaction.

Understanding Le Chatelier's principle helps chemists to control the conditions to favor the production of desired substances. It also aids in interpreting how the equilibrium constant of a reaction might change with temperature, since equilibrium constants are temperature-dependent.

Reaction Quotient

The reaction quotient (Q) helps us determine the direction in which a reaction is likely to proceed at any given moment. It is calculated using the same expression as the equilibrium constant, but with the current concentrations (or partial pressures) of the reactants and products, not necessarily those at equilibrium.

The reaction quotient is compared to the equilibrium constant (K) as follows:

• If \(Q > K\), there are too many products, and the reaction will proceed in the reverse direction.
• If \(Q < K\), there are too many reactants, and the reaction will go forward.
• If \(Q = K\), the system is at equilibrium, and no shift is needed.

This comparison enables chemists to predict how changes in a system will affect the position of equilibrium. For the given exercise, we analyze reaction (1) and (2), calculate their respective equilibrium constants, and use them to deduce the equilibrium constant for reaction (3), following the principles of reaction quotients and Gibbs free energy.