Problem 43 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Steam decomposes at high temperature according to the equation: \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) ; \quad \Delta H^{\circ}\) \(=240 \mathrm{~kJ} / \mathrm{mole}\) and \(\Delta S^{\circ}=50 \mathrm{JK}^{-1} / \mathrm{mole}\) The temperature at which the equilibrium constant \(\left(K_{\mathrm{p}}^{\circ}\right)\) becomes \(1.0\), is (a) \(4.8 \mathrm{~K}\) (b) \(4800 \mathrm{~K}\) (c) \(480 \mathrm{~K}\) (d) Impossible

Step-by-Step Solution

Answer

The temperature at which the equilibrium constant \(K_{\mathrm{p}}^\circ\) is 1.0 is 4800 K.

Step 1: Understand Gibbs Free Energy

The change in Gibbs free energy (\text{\text{\(\Delta G^\circ \)}}) for a reaction at equilibrium is zero. This allows us to set up the equation \text{\text{\(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0 \)}}, where \text{\text{\(\Delta H^\circ \)}} is the change in enthalpy, \text{\text{\(T\)}} is the temperature in Kelvin, and \text{\text{\(\Delta S^\circ \)}} is the change in entropy.

Step 2: Rearrange the Gibbs Free Energy Equation

To find the temperature at which the equilibrium constant \text{\text{\(K_\mathrm{p}^\circ\)}} is 1.0, we need to isolate \text{\text{\(T\)}} from the equation: \text{\text{\(T = \frac{\Delta H^\circ}{\Delta S^\circ}\)}}.

Step 3: Calculate the Temperature (T)

Substitute the given values into the rearranged Gibbs free energy equation: \text{\text{\(T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{240 \times 10^3 J/mol}{50 J/K \cdot mol} = 4800 K\)}}.

Step 4: Conclude with the Correct Answer

The temperature at which the equilibrium constant \text{\text{\(K_{\mathrm{p}}^\circ\)}} is 1.0 is 4800 K, which corresponds to answer choice (b).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in which the rate of the forward reaction is equal to the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time, not necessarily equal. This concept is a fundamental aspect of chemical reactions and can be observed in both simple and complex processes.

For example, when steam decomposes at high temperature to form hydrogen and oxygen gas, the system eventually reaches a point where as much water is decomposing into hydrogen and oxygen as is being formed from them—the system is at equilibrium.

This can be represented using a double arrow in the reaction equation:
\( 2 \text{H}_2\text{O}(g) \rightleftharpoons 2 \text{H}_2(g) + \text{O}_2(g) \).

To predict and understand the position of equilibrium, we use the equilibrium constant (\( K \)), which quantifies the ratio of the concentration of the products to the reactants, each raised to the power of its coefficient in the balanced equation.

Enthalpy Change

Enthalpy change (\( \text{Δ}H \)) is a measure of the total energy change in a system during a reaction, occurring at constant pressure. It represents the heat absorbed or released.

In reactions where energy is released, such as combustion, the change is negative (exothermic). Conversely, in reactions where energy is absorbed, like the decomposition of steam in the given exercise, the change is positive (endothermic).

The enthalpy change for the decomposition of steam to form hydrogen and oxygen gases is positive, illustrating that heat is absorbed from the surroundings:
\( \text{Δ}H^\text{o} = 240\text{ kJ/mol} \).

This energy change plays a critical role in determining whether a reaction is spontaneous under certain conditions and is closely related to the concept of Gibbs free energy.

Entropy Change

Entropy change (\( \text{Δ}S \)) refers to the change in entropy of a system during a reaction, which quantifies the degree of disorder or randomness.

In the universe, processes naturally move towards greater disorder, and the entropy of a closed system can never decrease.

In the reaction of steam decomposition, since gases are being formed from a liquid, there is an increase in entropy:
\( \text{Δ}S^\text{o} = 50\text{ J/K}\text{⋅mol} \).

The second law of thermodynamics states that for a process to be spontaneous, the total entropy of the universe (system plus surroundings) must increase. This concept integrates with Gibbs free energy to determine the spontaneity of reactions.

Equilibrium Constant

The equilibrium constant (\( K \)) is a number that expresses the relationship between the concentrations of the products and the reactants at chemical equilibrium. It is specific to a particular reaction at a given temperature.

Mathematically, for the reaction the equilibrium constant for the given example would be represented as:
\( K = \frac{[\text{H}_2]^2[\text{O}_2]}{[\text{H}_2\text{O}]^2} \), where square brackets indicate concentrations.

If \( K \) is much greater than 1, the products are favored at equilibrium; if \( K \) is much less than 1, the reactants are favored. When \( K=1 \), as in the exercise, it signifies that neither reactants nor products are favored, and the system is at a state of ‘perfect’ balance, though it doesn't mean their concentrations are equal. Gibbs free energy is related to the equilibrium constant by the equation \( \text{Δ}G^\text{o} = -RT\text{ln}K \), where \( R \) is the gas constant and \( T \) is the temperature.