Question

Calculate \(K_{\mathrm{eq}}\) for a reaction with \(\Delta G^{0}=-17.1 \mathrm{~kJ} / \mathrm{mol}(-4.09 \mathrm{kcal} / \mathrm{mol})\) at \(328 \mathrm{~K}\). Compare this value to the \(1 \times 10^{3}\) seen at \(298 \mathrm{~K}\).

Short answer

Answer: When the temperature is increased from 298 K to 328 K, the equilibrium constant increases from \(1 \times 10^3\) to approximately 5466. This indicates that the reaction is more favorable at higher temperatures, causing the equilibrium to shift towards the products.

Step-by-step solution

Extract the given information

We are given the following information: - \(\Delta G^0 = -17.1 \,\text{kJ/mol} = -17100 \,\text{J/mol}\) - Temperature, T = 328 K - Universal gas constant, R = 8.314 \(\text{J/mol}\text{.K}\) (it's important to have the same units as \(\Delta G^0\))

Rearrange the formula to solve for \(K_\text{eq}\)

We need to rearrange the equation \(\Delta G^0 = -RT\ln{K_\text{eq}}\) to solve for \(K_\text{eq}\): \(K_\text{eq} = e^{(-\Delta G^0)/(RT)}\)

Substitute the given values into the formula

Plug the given values of \(\Delta G^0\), R, and T into the formula to find \(K_\text{eq}\): \(K_\text{eq} = e^{(-(-17100 \,\text{J/mol}))/(8.314\,\text{J/mol.K} \times 328\,\text{K})}\)

Calculate the value of \(K_\text{eq}\) at 328 K

Now, we can calculate the equilibrium constant at 328 K: \(K_\text{eq} = e^{(17100)/(8.314 \times 328)} \approx 5466\)

Compare the value of \(K_\text{eq}\) at 328 K to the value at 298 K

We have found that the \(K_\text{eq}\) at 328 K is approximately 5466. The value given at 298 K is \(1 \times 10^3\). Comparing these values, we can see that the equilibrium constant at 328 K is higher than that at 298 K. This suggests that the reaction is more favorable at higher temperatures, thus shifting the equilibrium towards the products as temperature increases.

Key concepts

Gibbs Free Energy

Gibbs free energy, denoted as \(\Delta G\), is a thermodynamic quantity that represents the maximum reversible work that may be performed by a thermodynamic system at constant temperature and pressure. It's a pivotal concept in predicting the spontaneity of a process. A process is spontaneous if it leads to a decrease in the Gibbs free energy, typically indicated by a negative \(\Delta G\).

In the context of chemical reactions, the standard Gibbs free energy change \(\Delta G^0\) is particularly significant. It is calculated under standard conditions, which includes a temperature of 298 K and 1 atm pressure for each gas involved in the reaction. The equation \(\Delta G = \Delta G^0 + RT\ln Q\), where \(Q\) is the reaction quotient, connects the Gibbs free energy at any point in the reaction with the standard Gibbs free energy.

Reaction Temperature Effect

The effect of temperature on a reaction can be analyzed through the lens of Gibbs free energy. According to the relationship \(\Delta G = \Delta H - T\Delta S\), where \(\Delta H\) is the change in enthalpy and \(\Delta S\) is the change in entropy, we see that temperature \(T\) plays a crucial role.

For an exothermic reaction (negative \(\Delta H\)), an increase in temperature can make \(\Delta G\) more positive, thus making the reaction less spontaneous. Conversely, for an endothermic reaction (positive \(\Delta H\)), an increase in temperature can make \(\Delta G\) more negative, favoring the reaction's spontaneity. The solution provided illustrates how the equilibrium constant \(K_{eq}\) for a reaction can increase with temperature, showing a greater product formation at higher temperatures.

Chemical Equilibrium

Chemical equilibrium is the state of a reversible chemical reaction where the rates of the forward and reverse reactions are equal, leading to no net change in the concentrations of the reactants and products over time. The position of equilibrium is described by the equilibrium constant \(K_{eq}\), which is a ratio of the concentration of products to reactants, each raised to the power of their stoichiometric coefficients.

The equilibrium constant is tied to the Gibbs free energy change for the reaction by the relationship \(\Delta G^0 = -RT\ln K_{eq}\). As highlighted in the exercise, by calculating \(K_{eq}\) using the \(\Delta G^0\) value and temperature, we can determine the favorability of a reaction under certain conditions. A larger \(K_{eq}\) indicates that the products are favored at equilibrium, as seen in the comparison of equilibrium constants at different temperatures.

Thermodynamics in Chemistry

Thermodynamics is the branch of physical chemistry that deals with heat and temperature and their relation to energy, work, radiation, and properties of matter. It establishes general principles governing energy conversion and the directionality of physical and chemical processes.

The four laws of thermodynamics lay the groundwork, with the first law concerning conservation of energy, and the second highlighting the increase of entropy, a measure of randomness or disorder, in naturally occurring processes.

These principles guide the behavior of chemical reactions and dictate criteria like the Gibbs free energy change for spontaneity. The thermodynamic properties such as \(\Delta G\), \(\Delta H\), and \(\Delta S\) are invaluable in predicting the equilibrium position and the dependency of chemical reactions on temperature, as clearly demonstrated in the problem's solution where a reaction’s equilibrium constant changes with a shift in temperature.