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About 5 billion kilograms of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), are made each year. Benzene is used as a starting material for many other compounds and as a solvent (although it is also a carcinogen, and its use is restricted). One compound that can be made from benzene is cyclohexane, \(\mathrm{C}_{6} \mathrm{H}_{12}\) $$\begin{array}{c} \mathrm{C}_{6} \mathrm{H}_{6}(\ell)+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12}(\ell) \\ \Delta_{\tau} H^{\circ}=-206.7 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} ; \\\ \Delta_{\tau} \mathrm{S}^{\circ}=-361.5 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}-\mathrm{rxn} \end{array}$$ Is this reaction predicted to be product-favored at equilibrium at \(25^{\circ} \mathrm{C} ?\) Is the reaction enthalpy-or entropy-driven?

Short Answer

Expert verified
The reaction is product-favored and enthalpy-driven at 25°C.

Step by step solution

01

Identify the Reaction Parameters

We are given the enthalpy change \( \Delta H^{\circ} = -206.7 \text{ kJ/mol} \) and the entropy change \( \Delta S^{\circ} = -361.5 \text{ J/(K} \cdot \text{mol)} \). The temperature \( T \) is \( 25^{\circ} \text{C} \), which is equivalent to 298 K.
02

Convert Units

First, convert the entropy change from joules to kilojoules to match the units of enthalpy. \[ \Delta S^{\circ} = -361.5 \text{ J/(K} \cdot \text{mol)} = -0.3615 \text{ kJ/(K} \cdot \text{mol)} \].
03

Calculate Gibbs Free Energy Change

Use the Gibbs free energy equation: \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). Substitute the known values: \[ \Delta G^{\circ} = -206.7 \text{ kJ/mol} - (298 \text{ K})(-0.3615 \text{ kJ/(K} \cdot \text{mol)}) \].
04

Simplify the Equation

Calculate the product of temperature and entropy change: \[ T \Delta S^{\circ} = 298 \text{ K} \times -0.3615 \text{ kJ/(K} \cdot \text{mol)}) = -107.757 \text{ kJ/mol} \].
05

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

Substitute \( T \Delta S^{\circ} \) into the Gibbs free energy equation: \[ \Delta G^{\circ} = -206.7 \text{ kJ/mol} + 107.757 \text{ kJ/mol} = -98.943 \text{ kJ/mol} \].
06

Evaluate Reaction Favorability

Since \( \Delta G^{\circ} \) is negative, the reaction is product-favored at equilibrium at \( 25^{\circ} \text{C} \).
07

Determine Reaction Driver

The reaction is enthalpy-driven because \( \Delta H^{\circ} \) is negative and the magnitude of \( -\Delta H^{\circ} \) is greater than \( T \Delta S^{\circ} \), making the enthalpy change the significant contributor to \( \Delta G^{\circ} \).

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Key Concepts

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

Enthalpy
Enthalpy is a measurement of energy in a thermodynamic system. It represents the heat content of a system at constant pressure. For the reaction where benzene is converted to cyclohexane, the enthalpy change (\(\Delta H^{\circ} = -206.7 \text{ kJ/mol}\)) is crucial.
  • A negative enthalpy change indicates that the reaction releases heat, making it exothermic.
  • This means energy is released to the surroundings as the reaction occurs.
In our specific reaction, the substantial negative enthalpy suggests that the process often leans on enthalpic favorability, or being enthalpy-driven. This is a key driver in determining if a reaction may be product-favored at equilibrium, given that significant energy is released which can stabilize the system.
Entropy
Entropy measures the disorder or randomness of a system. It reflects how energy is distributed among particles. For the benzene to cyclohexane reaction, the entropy change (\(\Delta S^{\circ} = -361.5 \text{ J/(K} \cdot \text{mol)}\)) is negative.
  • A negative entropy change suggests that the system becomes more ordered during the reaction.
  • This means the formation of cyclohexane results in a decrease in randomness.
Though entropy isn't the main driver here (since the magnitude of\(T \Delta S^{\circ}\) is smaller than\(\Delta H^{\circ}\)), understanding its role is crucial. It hints at a decrease in possible arrangements of energy levels or configurations in the products compared to the reactants.
Reaction Equilibrium
Reaction equilibrium occurs when the rates of the forward and reverse reactions are equal. At this point, the concentrations of reactants and products remain constant. However, they are not necessarily equal!
  • A negative Gibbs Free Energy (\(\Delta G^{\circ} = -98.943 \text{ kJ/mol}\)) tells us the reaction is product-favored at a standard condition like 25°C.
  • This means at equilibrium, more cyclohexane than benzene and hydrogen will be present.
The balance of enthalpy and entropy factors achieves this equilibrium. For our reaction, it's primarily the release of heat, indicative of being enthalpy-driven, that tips the balance in favor of product formation at equilibrium.

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Most popular questions from this chapter

The standard free energy change, \(\Delta_{\mathrm{r}} G^{\circ},\) for the formation of \(\mathrm{O}_{3}(\mathrm{g})\) from \(\mathrm{O}_{2}(\mathrm{g})\) is \(+163.2 \mathrm{kJ} / \mathrm{mol}\) -ren at \(25^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{p}}\) at this temperature for the equilibrium $$3 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}_{3}(\mathrm{g})$$ Comment on the sign of \(\Delta_{r} G^{\circ}\) and the magnitude of \(K_{\mathrm{p}}.\)

Determine if each of the following statements is true or false (a) For a reaction that can yield more than one product, the most stable product is always formed in the greatest amount. (b) For a reaction that can yield more than one product, the pathway that has the lowest activation energy always occurs more quickly. (c) Reactions that are thermodynamically favored are always fast. (d) Diamond is always more stable than graphite.

The enthalpy of vaporization of liquid diethyl ether, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O},\) is \(26.0 \mathrm{kJ} / \mathrm{mol}\) at the boiling point of \(35.0^{\circ} \mathrm{C} .\) Calculate \(\Delta S^{\circ}\) for a vapor-to-liquid transformation at \(35.0^{\circ} \mathrm{C}.\)

Identify the following processes as either spontaneous or not spontaneous. (a) Liquid water turns to ice when placed in a freezer at \(-5^{\circ} \mathrm{C}\) (b) Nitrogen gas is compressed to one half its original volume. (c) Sodium reacts with water forming \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{NaOH}(\mathrm{aq})\) (d) Slightly soluble \(\operatorname{CaSO}_{4}\left(K_{\mathrm{sp}}=4.5 \times 10^{-5}\right)\) dis- solves in water to form a saturated solution.

Predict whether each of the following processes results in an increase in entropy in the system. (Define reactants and products as the system.) (a) Water vapor condenses to liquid water at \(90^{\circ} \mathrm{C}\) and 1 atm pressure. (b) The exothermic reaction of \(\mathrm{Na}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{g})\) forms \(\mathrm{NaCl}(\mathrm{s})\) (c) The endothermic reaction of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) produces an equilibrium mixture of \(\mathrm{H}_{2}(\mathrm{g})\) \(\mathrm{I}_{2}(\mathrm{g}),\) and \(\mathrm{HI}(\mathrm{g})\) (d) Solid NaCl dissolves in water forming a saturated solution.

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