Question

When permanganate ions in aqueous solution react with cobalt metal in strong acid, the equation for the reaction that takes place is $$ \begin{array}{r} 2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q)+5 \mathrm{Co}(s) \longrightarrow \\ 2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{Co}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l) \\\ \Delta H^{\circ}=-2024.6 \mathrm{~kJ} ; \Delta G^{\circ} \text { at } 25^{\circ} \mathrm{C}=-1750.9 \mathrm{~kJ} \end{array} $$ (a) Calculate \(\Delta S^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\). (b) Calculate \(S^{\circ}\) for \(\mathrm{Co}^{2+}\), given \(S^{\circ}\) for Co is \(30.04 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\).

Short answer

Also, find the standard entropy (S°) for \(Co^{2+}(aq)\), given the standard entropy for \(Co(s)\). The standard entropy change for the reaction at 25°C is found to be -0.918 kJ/mol K or -918 J/mol K. The standard entropy for \(Co^{2+}(aq)\) is calculated to be -153.76 J/mol K.

Step-by-step solution

Convert the temperature to Kelvin

To begin, we need to convert the given temperature, 25°C, to Kelvin. Remember that: $$ K = °C + 273.15 $$ Therefore: $$ 25°C + 273.15 = 298.15K $$ The temperature is 298.15K.

Calculate ∆S° using the Gibbs-Helmholtz Equation

Now that we have the temperature in Kelvin, we can use the Gibbs-Helmholtz Equation to find the standard entropy change for the reaction. The equation is as follows: $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} $$ Rearranging the equation to solve for ∆S°: $$ \Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} $$ Substituting the given values: $$ \Delta S^{\circ} = \frac{-2024.6 \text{ kJ/mol} - (-1750.9 \text{ kJ/mol})}{298.15K} $$ Calculating the entropy change: $$ \Delta S^{\circ} = \frac{-273.7 \text{ kJ/mol}}{298.15K} = -0.918 \text{ kJ/mol K} $$ The standard entropy change for the reaction is -0.918 kJ/mol K or -918 J/mol K.

Calculate the standard entropy for \(Co^{2+}\)

Finally, we can calculate the standard entropy for \(Co^{2+}(aq)\). We are given the standard entropy for \(Co(s)\), and we have found the standard entropy change for the reaction. Using the stoichiometry of the balanced chemical equation, we can determine the entropy of \(Co^{2+}(aq)\). For every 5 moles of \(Co(s)\), there are 5 moles of \(Co^{2+}(aq)\) produced. Thus: $$ \Delta S^{\circ}_{reaction} = 5 \times (S^{\circ}_{Co^{2+}} - S^{\circ}_{Co}) $$ Rearrange the equation to solve for \(S^{\circ}_{Co^{2+}}\): $$ S^{\circ}_{Co^{2+}} = \frac{\Delta S^{\circ}_{reaction}}{5} + S^{\circ}_{Co} $$ Substitute the values: $$ S^{\circ}_{Co^{2+}} = \frac{-918 \text{ J/mol K}}{5} + 30.04 \text{ J/mol K} = -153.76 \text{ J/mol K} $$ The standard entropy for \(Co^{2+}(aq)\) is -153.76 J/mol K.

Key concepts

Gibbs-Helmholtz Equation

Understanding the Gibbs-Helmholtz equation is crucial in the world of thermodynamics, especially when studying chemical reactions. It's a mathematically elegant tool that relates the change in Gibbs free energy (\f\(\f\backslashDelta G^{\f\backslashcirc}\f\)) of a system to its enthalpy change (\f\(\f\backslashDelta H^{\f\backslashcirc}\f\)) and entropy change (\f\(\f\backslashDelta S^{\f\backslashcirc}\f\)) at a constant temperature.

The equation is commonly written as:
\f[\f\backslashDelta G^{\f\backslashcirc} = \f\backslashDelta H^{\f\backslashcirc} - T\f\backslashDelta S^{\f\backslashcirc}\f]
Here, \f\(T\f\) is the temperature in Kelvin, emphasizing the need for temperature consistency across thermodynamic calculations. When solving problems, remember to convert temperatures from Celsius to Kelvin by adding 273.15, ensuring that you're working in the correct temperature scale for thermodynamic equations.

By rearranging the equation, it can also be used to solve for the standard entropy change (\f\(\f\backslashDelta S^{\f\backslashcirc}\f\)) when \f\(\f\backslashDelta G^{\f\backslashcirc}\f\) and \f\(\f\backslashDelta H^{\f\backslashcirc}\f\) are known, which is often necessary when working with reactions and processes where direct entropy measurements are challenging to obtain.

Standard Entropy Change

The concept of standard entropy change (\f\(\f\backslashDelta S^{\f\backslashcirc}\f\)) is essential for grasping the behavior of a system during a chemical reaction. Entropy, often described as a measure of disorder, is a fundamental principle in chemistry that helps predict the spontaneity of a process. The standard entropy change refers to the entropy difference between products and reactants under standard conditions (usually 1 bar of pressure and a specified temperature, typically 25°C or 298.15K).

When the entropy of the system increases during a reaction, \f\(\f\backslashDelta S^{\f\backslashcirc}\f\) is positive, indicating a more disordered system. Conversely, a negative \f\(\f\backslashDelta S^{\f\backslashcirc}\f\) suggests a decrease in disorder. For the permanganate and cobalt reaction presented in the exercise, the standard entropy change is calculated using the Gibbs-Helmholtz equation, demonstrating a decrease in entropy.

Understanding how to compute \f\(\f\backslashDelta S^{\f\backslashcirc}\f\) fosters a deeper comprehension of how energy is redistributed during chemical transformations and is vital in predicting the direction and extent of chemical reactions.

Enthalpy Change

Another critical thermodynamic concept is the enthalpy change (\f\(\f\backslashDelta H^{\f\backslashcirc}\f\)), which represents the heat absorbed or released by the system during a reaction at constant pressure. The enthalpy change provides insight into the energy dynamics of the reaction. If \f\(\f\backslashDelta H^{\f\backslashcirc}\f\) is negative, the process is exothermic, releasing heat to the surroundings. On the flip side, a positive \f\(\f\backslashDelta H^{\f\backslashcirc}\f\) indicates an endothermic reaction where the system absorbs heat from its environment.

In the provided exercise, the enthalpy change is significantly negative, thus the reaction between permanganate ions and cobalt metal is highly exothermic. This strong release of energy often favors the formation of products, but it's only part of the story. By combining this information with entropy data, we can get a complete picture of the reaction's spontaneity through the Gibbs free energy. It's essential to not only calculate but also to interpret \f\(\f\backslashDelta H^{\f\backslashcirc}\f\) in the context of entropy and Gibbs free energy to fully understand chemical thermodynamics.