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Calculate \(\Delta \mathrm{G}^{\circ}\) for the reaction below and tell whether it is spontaneous or non-spontaneous under standard conditions at \(25^{\circ} \mathrm{C}\). \(2 \mathrm{~S}(\mathrm{~s})+3 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{4}(l)\) \(\Delta \mathrm{H}^{\circ}=-1056 \mathrm{~kJ} / \mathrm{mol}\) \(\Delta \mathrm{S}^{0}=-505 \mathrm{~J} / \mathrm{mol}\) a. \(\Delta \mathrm{G}^{\circ}=-906 \mathrm{~kJ}\) and the process is spontaneous b. \(\Delta \mathrm{G}^{0}=-1206\) and the process is spontaneous c. \(\Delta \mathrm{G}^{\circ}=-906 \mathrm{~kJ}\) and the process is non- spontaneous d. \(\Delta \mathrm{G}^{0}=-1206 \mathrm{~kJ}\) and the process is non- spontaneous

Short Answer

Expert verified
The answer is (a): \(\Delta G^{\circ} = -906 \, \text{kJ}\) and the process is spontaneous.

Step by step solution

01

Identify Given Data

We are given \(\Delta H^{\circ} = -1056 \, \text{kJ/mol}\) and \(\Delta S^{\circ} = -505 \, \text{J/mol K}\). Note that \(\Delta S^{\circ}\) is in Joules, while \(\Delta H^{\circ}\) is in kilojoules.
02

Convert Units if Necessary

Since \(\Delta S^{\circ}\) is given in \(\text{J/mol K}\), convert it to \(\text{kJ/mol K}\) by dividing by 1000: \(\Delta S^{\circ} = -0.505 \, \text{kJ/mol K}\).
03

Apply Gibbs Free Energy Formula

Use the formula for Gibbs free energy change \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\). Here, \(T = 298 \, \text{K}\).
04

Calculate \(\Delta G^{\circ}\)

Substitute the known values into the equation: \[\Delta G^{\circ} = -1056 \, \text{kJ/mol} - (298 \, \text{K} \times -0.505 \, \text{kJ/mol K})\].
05

Perform the Calculation

Calculate the change in temperature and entropy: \(298 \, \text{K} \times -0.505 \, \text{kJ/mol K} = -150.49 \, \text{kJ/mol}\).
06

Finalize \(\Delta G^{\circ}\) Value

Insert this into the initial equation: \[\Delta G^{\circ} = -1056 \, \text{kJ/mol} + 150.49 \, \text{kJ/mol} = -905.51 \, \text{kJ/mol}\]. Since the calculation rounds to -906 \(\text{kJ}\), choose the closest answer: \(-906 \, \text{kJ}\).
07

Determine Spontaneity

A negative \(\Delta G^{\circ}\) value indicates that the reaction is spontaneous under standard conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spontaneous Reactions
In chemistry, a spontaneous reaction is one that occurs naturally without any external help. It takes place because the energy state of the products is lower than that of the reactants. To determine if a reaction is spontaneous, Gibbs Free Energy ( \(\Delta G\)) is a crucial factor.
  • If \(\Delta G\) is negative, the reaction is spontaneous at the specific conditions, meaning it can proceed on its own.
  • If \(\Delta G\) is positive, the reaction is non-spontaneous, requiring energy input to occur.
For the given chemical reaction, the calculation resulted in \(\Delta G^{\circ} = -906 \, \text{kJ/mol}\), which clearly indicates it is a spontaneous reaction under standard conditions. The energy barrier is effectively overcome, leading the reaction to proceed naturally towards product formation.
Enthalpy
Enthalpy, represented as \(\Delta H\), reflects the total heat content of a system. It tells us about the heat absorbed or released during a reaction at constant pressure. A reaction might be:
  • Exothermic: When \(\Delta H\) is negative, it releases heat into its environment, making the surroundings warmer. This was seen in the example where \(\Delta H^{\circ} = -1056 \, \text{kJ/mol}\).
  • Endothermic: When \(\Delta H\) is positive, it absorbs heat, making the surroundings cooler.
In the original exercise, we observed an exothermic reaction, releasing a significant amount of energy as heat, making it more favorable for a spontaneous nature. This large release of energy is often a strong driving force for the spontaneity of reactions.
Entropy
Entropy, denoted as \(\Delta S\), measures a system’s degree of disorder or randomness. Generally, systems tend toward higher entropy, meaning more disorder is typically more favorable.
  • Positive \(\Delta S\) corresponds to increasing disorder, often pushing toward spontaneity.
  • Negative \(\Delta S\) suggests decreasing disorder or the system becoming more ordered.
In the provided problem, \(\Delta S^{\circ} = -505 \, \text{J/mol K}\), indicating the reaction results in decreased disorder. Even though the entropy change is negative, the overall spontaneity is achieved thanks to the large negative \(\Delta H\), which outweighs the ordering effect, leading to a negative \(\Delta G^{\circ}\).
Standard Conditions
Standard conditions refer to a set of specific environmental settings for which reactions are measured, usually set at 25°C (298 K) and 1 atmosphere pressure. Under these conditions, it becomes easier to compare and predict behaviors of different chemical reactions.
In controlled environments like these, properties such as enthalpy and entropy are often marked with a superscript "\(\circ\)" to denote that they are calculated under standard conditions.For this exercise, the determination of spontaneity involving \(\Delta G^{\circ}\) took place under these standard circumstances, making it straightforward to predict the reaction’s tendency to proceed without additional interventions. By understanding these conditions, one can reliably judge how a system behaves under a universal reference frame.

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Most popular questions from this chapter

The lattice energy of \(\mathrm{NaCl}\) is \(780 \mathrm{~kJ} \mathrm{~mol}^{-1}\) The enthalpies of hydration of \(\mathrm{Na}^{+}(\mathrm{g})\) and \(\mathrm{Cl}^{-}(\mathrm{g})\) ions are \(-406\) \(\mathrm{kJ} \mathrm{mol}^{-1}\) and \(-364 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The enthalpy of solution of \(\mathrm{NaCl}(\mathrm{s})\) is a. \(10 \mathrm{~kJ} \mathrm{~mol}^{-1}\) b. \(-10 \mathrm{~kJ} \mathrm{~mol}^{-1}\) c. \(736 \mathrm{~kJ} \mathrm{~mol}^{-1}\) d. \(100 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

In thermodynamics, a process is called reversible when a. surroundings and system change into each other b. there is no boundary between system and surroundings c. the surroundings are always in equilibrium with the system d. the system changes into the surroundings spontaneously [IIT 2001]

Bond dissociation enthalpy of \(\mathrm{H}_{2}, \mathrm{Cl}_{2}\) and \(\mathrm{HCl}\) are 434, 242 and \(431 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. Enthalpy of formation of \(\mathrm{HCl}\) is a. \(-93 \mathrm{~kJ} \mathrm{~mol}^{-1}\) b. \(245 \mathrm{~kJ} \mathrm{~mol}^{-1}\) c. \(93 \mathrm{~kJ} \mathrm{~mol}^{-1}\) d. \(-245 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

For which of the following reaction is the standard entropy of reaction \(\Delta \mathrm{S}^{\circ}\) positive? a. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) c. \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) d. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

(A): Enthalpy changes are positive when \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). \(10 \mathrm{H}_{2} \mathrm{O}, \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and salts like \(\mathrm{NaCl}, \mathrm{KCl}\) etc. which do not form hydrates are dissolved in water. But enthalpy changes are negative when anhydrous salts capable of forming hydrates are dissolved in water. (R): The difference in the behaviour is due to large differences in the molecular weights of hydrated and anhydrous salts. The substances with larger molecular weights usually show positive enthalpy change on dissolution.

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