Problem 73 - Chemistry An Atoms First Approach

Question

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: $$\begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6} \text { at } 600 . \mathrm{K}\\\ &2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37} \mathrm{at} 600 . \mathrm{K} \end{aligned}$$

Step-by-Step Solution

Answer

The standard Gibbs free energy change for the reaction \(\mathrm{H}_{2}\mathrm{O}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}_{2}(g)\) at \(600\, \mathrm{K}\) is approximately \(-14,109\, \text{J/mol}\).

Step 1: Related reactions and equilibrium constants

We have the following related reactions and their equilibrium constants at \(600\, \mathrm{K}\): 1. \(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}_{2}(g)\) with \(K_1 = 2.3 \times 10^{6}\) 2. \(2\, \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2\, \mathrm{H}_{2}\mathrm{O}(g)\) with \(K_2 = 1.8 \times 10^{37}\) We want to find the equilibrium constant for the reaction: \(\mathrm{H}_{2}\mathrm{O}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}_{2}(g)\)

Step 2: Find the desired reaction and equilibrium constant

We can manipulate the related reactions to get the desired reaction and its equilibrium constant. From reaction 2, we will make some modifications: $$\frac{1}{2}(2\, \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2\, \mathrm{H}_{2}\mathrm{O}(g))$$ Which gives us the following reaction: $$\mathrm{H}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}(g)$$ For a reaction multiplied by a factor, the new equilibrium constant is the old one raised to the power of that factor, so we will calculate the new equilibrium constant as: $$K'_2 = (K_2)^{\frac{1}{2}} = (1.8 \times 10^{37})^{\frac{1}{2}}$$ Now, we can combine Reaction 1 and the modified Reaction 2 to get the desired reaction: $$ (\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}_{2}(g)) + (\mathrm{H}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}(g)) $$ $$\Rightarrow \mathrm{H}_{2}\mathrm{O}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}_{2}(g)$$ We'll call the equilibrium constant for this reaction \(K'\) and calculate it by multiplying the equilibrium constants of the related reactions: $$K' = K_1 \times K'_2$$

Step 3: Calculate the standard Gibbs free energy change

Once we have the equilibrium constant for the desired reaction, we can calculate the standard Gibbs free energy change, \(\Delta G^{\circ}\). The formula connecting the equilibrium constant and Gibbs free energy change is: $$\Delta G^{\circ} = -RT\ln{K'}$$ where \(R\) is the gas constant \((8.314\, \text{J/mol}\cdot\text{K})\) and \(T\) is the temperature \((600\, \mathrm{K})\). Plugging in the values, we have: $$\Delta G^{\circ} = -8.314\, \text{J/mol}\cdot\text{K} \cdot 600\, \mathrm{K} \cdot \ln{(K')}$$ Now, we have all the equations we need to solve for \(\Delta G^{\circ}\).

Step 4: Calculate the answer

First, calculate \(K'_2\): $$K'_2 = (1.8 \times 10^{37})^{\frac{1}{2}} = 4.243\times10^{18}$$ Next, calculate \(K'\): $$K' = K_1 \times K'_2 = 2.3\times 10^6 \times 4.243\times10^{18} = 9.758\times10^{24}$$ Finally, calculate \(\Delta G^{\circ}\): $$\Delta G^{\circ} = -8.314\, \text{J/mol}\cdot\text{K} \cdot 600\, \mathrm{K} \cdot \ln{(9.758\times10^{24})}$$ $$\Delta G^{\circ} \approx -14,109\, \text{J/mol}$$ So, the standard Gibbs free energy change for the given reaction at \(600\, \mathrm{K}\) is approximately \(-14,109\, \text{J/mol}\).

Key Concepts

Equilibrium Constant

The equilibrium constant, represented by the symbol \( K \), is a fundamental concept in chemical reactions. It measures the ratio of the concentrations of products to reactants at equilibrium. The main idea is that, at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. The equilibrium constant thus provides a snapshot of the concentration ratios at equilibrium. For any given reaction at a constant temperature, the value of the equilibrium constant remains consistent. This makes it a very useful tool to predict the direction of a reaction based on the initial concentrations of reactants and products.

• A large \( K \) means that a reaction favors the formation of products, implying the reaction proceeds almost to completion.
• A small \( K \) indicates that the reaction favors the formation of reactants, meaning not much product is formed at equilibrium.
• \( K \) is dimensionless and is influenced by temperature.

Understanding the equilibrium constant is crucial for calculating the Gibbs free energy change, as seen in our exercise, which determines how spontaneously a reaction occurs.

Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that deals with the study of energy changes, particularly with heat and work, during chemical reactions. It delves deeply into the concepts of enthalpy, entropy, and Gibbs free energy.
**Key Concepts:**

• **Enthalpy (\( \Delta H \))**: Measures total heat content in a system. It indicates whether a reaction is exothermic (heat releasing) or endothermic (heat absorbing).
• **Entropy (\( \Delta S \))**: Represents the disorder or randomness of a system. An increase in entropy usually indicates an increase in disorder.
• **Gibbs Free Energy (\( \Delta G \))**: Combines enthalpy and entropy into a single number that can predict the spontaneity of a reaction. The equation is: \( \Delta G = \Delta H - T \Delta S \), where \( T \) is the temperature in Kelvin.

In the context of our problem, we focus on the Gibbs free energy equation relating it to equilibrium constants: \( \Delta G^{\circ} = -RT\ln K \). This reveals how the tendency of a reaction can be inferred from thermodynamic principles.

Standard Conditions

Standard conditions are a set of pre-defined values that are typically used to standardize thermodynamic calculations. These conditions allow scientists and engineers to compare different reactions under a uniform benchmark.
**Standard Conditions Typically Include:**

• Pressure of \( 1 \) atmosphere.
• Concentration of \( 1 \) mol/L for solutions.
• A baseline temperature, often \( 298.15 \) K \((25^\circ C)\).

In practice, physiological environments or laboratory conditions might differ, but these standard conditions provide a consistent starting point for theoretical calculations. In our exercise, however, the reaction is evaluated at a non-standard temperature of \( 600 \, \mathrm{K} \). This deviation is significant for calculating Gibbs free energy and necessitates the use of the specific temperature provided when applying the equation \( \Delta G^{\circ} = -RT\ln K \). As temperature can significantly impact Gibbs free energy and equilibrium constants, understanding standard and non-standard conditions ensures accurate predictions in different scenarios.