Problem 126 - Chemistry: An Atoms-Focused Approach

Question

Making Hydrogen Debate continues on the practicality of using \(\mathrm{H}_{2}\) gas as a fucl for cars. The equilibrium constant \(K_{e}\) for the reaction $$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)$$ is \(1.0 \times 10^{5}\) at \(25^{\circ} \mathrm{C} .\) Starting with this value, calculate the value of \(\Delta G_{r r n}^{\infty}\) at \(25^{\circ} \mathrm{C}\) and, without doing any calculations, guess the sign of \(\Delta H_{\text {rent }}\)

Step-by-Step Solution

Answer

Question: Calculate the standard Gibbs free energy change (\(\Delta G_{r r n}^{\infty}\)) for the reaction between carbon monoxide (CO) and water (H2O) to produce carbon dioxide (CO2) and hydrogen gas (H2) with an equilibrium constant \(K_e = 1.0 \times 10^{5}\) at 25°C, and guess the sign of the standard enthalpy change (\(\Delta H_{\text {rent}}\)) without doing any calculations. Answer: The standard Gibbs free energy change (\(\Delta G_{r r n}^{\infty}\)) for the given reaction is -1.94 x 10^5 J/mol, and we can guess that the sign of the standard enthalpy change (\(\Delta H_{\text {rent}}\)) is negative, indicating that the reaction is exothermic.

Step 1: Recall the relationship between \(\Delta G_{r r n}^{\infty}\) and \(K_e\)

Using the formula \(\Delta G_{r r n}^{\infty} = -RT\ln{K_e}\), we have the relationship between the standard Gibbs free energy change and the equilibrium constant. In this formula, \(R\) is the gas constant (8.31 J/mol K) and \(T\) is the temperature in Kelvin.

Step 2: Convert the temperature to Kelvin

First, convert the given temperature from Celsius to Kelvin: \(T = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \mathrm{K}\)

Step 3: Calculate \(\Delta G_{r r n}^{\infty}\)

Now, plug the given values of \(K_e\), \(R\), and \(T\) into the equation we found in Step 1 to calculate \(\Delta G_{r r n}^{\infty}\): \(\Delta G_{r r n}^{\infty} = -RT\ln{K_e} = -(8.31 \frac{\mathrm{J}}{\mathrm{mol} \mathrm{K}})(298.15 \mathrm{K})\ln{(1.0 \times 10^{5})}\) Now, use a calculator to find the numerical value: \(\Delta G_{r r n}^{\infty} = -1.94 \times 10^{5} \frac{\mathrm{J}}{\mathrm{mol}}\)

Step 4: Guess the sign of \(\Delta H_{\text {rent}}\)

Since \(\Delta G_{r r n}^{\infty}\) is negative and we expect the reaction to be spontaneous, we can infer that the sign of \(\Delta H_{\text {rent}}\) is also negative. A negative value for \(\Delta H_{\text {rent}}\) indicates that the reaction is exothermic, which means that heat is released during the reaction. This outcome is reasonable because combustion reactions generally release heat.

Key Concepts

Equilibrium Constant

To understand Gibbs Free Energy better, it helps to first learn about the equilibrium constant, denoted as \(K_e\). The equilibrium constant is a number that indicates the ratio of product concentrations to reactant concentrations in a chemical reaction at equilibrium. This means that even though the reaction is still happening, the amounts of reactants and products remain constant.
- When \(K_e\) is much larger than 1, as in our exercise with a value of \(1.0 \times 10^{5}\), it suggests that at equilibrium, the products are much favored over the reactants.
- Conversely, if \(K_e\) is much smaller than 1, the reactants are favored.
In our exercise, knowing that \(K_e\) is large tells us that the reaction heavily favors the production of hydrogen, making it particularly relevant for fueling applications.

Exothermic Reaction

Exothermic reactions are processes that release energy in the form of heat to the surroundings. This means, during the reaction, energy stored in the chemical bonds of the reactants is released, generally increasing the temperature of the environment.
- A negative \(\Delta H_{\text{rent}}\), or change in enthalpy, signifies an exothermic process. In the context of our exercise, the negative \(\Delta G_{rrn}^{\infty}\) clues us in that \(\Delta H_{\text{rent}}\) may also be negative. This suggests that the reaction is releasing energy as it forms its products.
- Many combustion reactions are exothermic, making them efficient sources of energy.
Knowing the nature of the reaction helps anticipate energy dynamics, which is essential in applications like automotive fuel.

Temperature Conversion

In chemistry, temperatures are most often expressed in Kelvin, especially when using equations that involve physical constants, such as the gas constant \(R\). This is important because Kelvin, unlike Celsius or Fahrenheit, is an absolute temperature scale.
- Converting from Celsius to Kelvin is straightforward: simply add 273.15 to the Celsius temperature.
- For instance, our calculation began by converting 25°C to Kelvin. Thus, we get \(298.15\, \text{K}\).
Temperature affects reaction rates and equilibrium constants, so using Kelvin allows for a consistent approach in thermodynamic calculations. Understanding how to convert and apply temperature properly is crucial for accurately assessing reactions like the production of hydrogen gas.