Question

Consider the reaction: $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=-92.50 \mathrm{~kJ}\). Which of the following statements is(are) true? a. This is an endothermic reaction. b. \(\Delta S^{\circ}\) for this reaction is negative. c. If the temperature is increased, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will increase. d. \(\Delta G^{\circ}\) for this reaction has to be negative at all temperatures. e. When \(\Delta G^{\circ}\) for this reaction is negative, then \(K_{\mathrm{p}}\) is greater than \(1.00\).

Short answer

Only statement e is true: When $\Delta G^{\circ}$ for this reaction is negative, then $K_{\mathrm{p}}$ is greater than $1.00$.

Step-by-step solution

Statement a

We are given that \(\Delta H^{\circ} = -92.50 \ \text{kJ}\). Since it is negative, this reaction is exothermic, not endothermic. So, statement a is false.

Statement b

To assess the sign of \(\Delta S^{\circ}\), we need to look at the reaction. Since the number of moles of reactants is equal to the number of moles of the product, it is not possible to definitively state that the entropy change \(\Delta S^{\circ}\) for this reaction is negative. Hence, statement b is not necessarily true.

Statement c

For this statement, we have to consider the effect of temperature on the equilibrium of the reaction using the Van't Hoff equation: $$\frac{d \ln K}{dT} = \frac{\Delta H^{\circ}}{RT^2}$$ Since the reaction is exothermic (\(\Delta H^{\circ}<0\)), increasing the temperature will shift the equilibrium towards the reactants (\(\mathrm{PCl_{3}}\) and \(\mathrm{Cl_{2}}\)). Consequently, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will decrease, not increase. So, statement c is false.

Statement d

The following relationship governs the standard Gibbs free energy change, \(\Delta G^{\circ}\): $$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$$ The statement claims that \(\Delta G^{\circ}\) is negative at all temperatures. However, without knowing the sign and magnitude of \(\Delta S^{\circ}\), we cannot conclusively determine the sign of \(\Delta G^{\circ}\) at all temperatures. Thus, statement d is not necessarily true.

Statement e

The relationship between \(\Delta G^{\circ}\) and the equilibrium constant \(K_\text{p}\) is: $$\Delta G^{\circ} = -RT \ln K_\text{p}$$ When \(\Delta G^{\circ}<0\), the ln term must be positive, which implies that \(K_\text{p}>1\). So, when \(\Delta G^{\circ}\) for this reaction is negative, \(K_{\mathrm{p}}\) is greater than \(1.00\). Thus, statement e is true. Summing up, only statement e is true.

Key concepts

Thermodynamics

Thermodynamics is the branch of physics dealing with heat and temperature and their relation to energy and work. It describes how thermal energy is converted into or from other forms of energy and how it affects matter. In chemistry, thermodynamics helps predict whether a reaction is spontaneous and the direction in which a reaction will proceed.

One of the core principles is the first law of thermodynamics, which states that energy cannot be created or destroyed, only transformed. This principle is particularly useful when analyzing chemical reactions, as it allows us to track energy changes in the form of heat or work.

• Reactions can be studied under constant pressure or volume, and the thermodynamic properties help determine the reaction behavior.
• The understanding of endothermic (heat-absorbing) and exothermic (heat-releasing) reactions is crucial.

In the context of the given exercise, the use of the standard enthalpy change, \(\Delta H^{\circ}\), is an essential component of thermodynamic analysis.

Enthalpy

Enthalpy (\(H\)) is a measurement of energy in a thermodynamic system. It is a state function and reflects the heat content of a system at constant pressure. The change in enthalpy, \(\Delta H\), during a reaction indicates whether a reaction is endothermic or exothermic.

For an exothermic reaction, where heat is released, \(\Delta H\) is negative. This is because the system loses energy to the surroundings. In an endothermic reaction, \(\Delta H\) is positive, as energy is absorbed from the surroundings.

• The reaction \(\text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g)\) has \(\Delta H^{\circ} = -92.50 \, \text{kJ}\), indicating it is exothermic.
• This tells us the forward reaction releases energy, making it energetically favorable under standard conditions.

Understanding enthalpy is crucial because it helps in predicting the heat flow for reactions and the position of equilibrium with changing temperature.

Entropy

Entropy (\(S\)) represents the randomness or disorder in a system. It is a key concept in determining how energy disperses in a system and is fundamental to the second law of thermodynamics, which states that the total entropy of an isolated system always increases over time.

Entropy change (\(\Delta S\)) during a reaction considers the difference in entropy between products and reactants. Reactions often proceed towards states that have higher entropy.

• In this exercise, predicting the exact sign of \(\Delta S^{\circ}\) isn't straightforward without further information, because the number of moles of gaseous reactants and products remain the same.
• However, considering changes in molecular complexity or phase can also influence \(\Delta S\).

Assessing entropy helps explain why certain reactions are spontaneous, even if they might be endothermic, as energy dispersal becomes a factor.

Gibbs Free Energy

Gibbs free energy (\(G\)) defines the maximum reversible work that can be performed by a thermodynamic system at constant temperature and pressure. It combines enthalpy and entropy into a single value, providing insight into the spontaneity of a process.
The formula is \(\Delta G = \Delta H - T\Delta S\). If \(\Delta G\) is negative, the reaction is spontaneous as more energy is released than required. Positive \(\Delta G\) signifies non-spontaneity and requires added energy to proceed.

• The exercise equation \(\Delta G^{\circ} = -RT \ln K_p\) links Gibbs energy to equilibrium constants.
• A negative \(\Delta G^{\circ}\) indicates \(K_p > 1\), meaning the reaction favors product formation.

Understanding Gibbs free energy helps chemists control reactions by adjusting conditions, like temperature, to achieve desired outcomes.