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The enthalpy and entropy change for the reaction \(\mathrm{Br}_{2}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{BrCl}(\mathrm{g})\) are \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(105 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) respectively. The temperature at which the reaction will be in equilibrium is (a) \(450 \mathrm{~K}\) (b) \(300 \mathrm{~K}\) (c) \(285.7 \mathrm{~K}\) (d) \(273 \mathrm{~K}\)

Short Answer

Expert verified
(c) 285.7 K

Step by step solution

01

Understand the Problem

We are given the enthalpy change (\(\Delta H\)) and the entropy change (\(\Delta S\)) for a reaction and are asked to find the temperature at which the reaction is in equilibrium. At equilibrium, the Gibbs free energy change (\(\Delta G\)) is zero.
02

Use the Gibbs Free Energy Equation

The formula for the Gibbs Free Energy is \(\Delta G = \Delta H - T \Delta S\). At equilibrium, \(\Delta G = 0\). So, the equation becomes \(0 = \Delta H - T \Delta S\). This can be rearranged to find \(T\), giving \(T = \frac{\Delta H}{\Delta S}\).
03

Substitute Given Values

Substitute the given values into the equation. \(\Delta H\) is given as \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\), which needs to be converted into \(\mathrm{~J} \mathrm{~mol}^{-1}\) (i.e., \(30000 \mathrm{~J} \mathrm{~mol}^{-1}\)). \(\Delta S\) is \(105 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\). So, \(T = \frac{30000}{105}\).
04

Calculate the Temperature

Perform the division to calculate \(T\). \(T = \frac{30000}{105} \approx 285.71 \mathrm{~K}\).
05

Choose the Closest Answer

Compare the calculated temperature with the provided options: (a) 450 K, (b) 300 K, (c) 285.7 K, (d) 273 K. The closest answer to 285.71 K is \(c) 285.7 \mathrm{~K}\).

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Key Concepts

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

Enthalpy Change
Enthalpy change is a measure of the total heat content in a system under constant pressure. It is symbolized as \( \Delta H \) and is typically expressed in units of joules per mole. In a reaction, enthalpy change indicates whether heat is absorbed or released. Enthalpy changes may be either positive or negative:
  • A positive \( \Delta H \) means the reaction absorbs heat (endothermic).
  • A negative \( \Delta H \) signifies that the reaction releases heat (exothermic).
In our exercise, the given \( \Delta H \) is \( 30 \, \text{kJ} \, \text{mol}^{-1} \), which is a positive value. This tells us that the reaction \( \mathrm{Br}_{2}(\ell)+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{BrCl}(\mathrm{g}) \)requires the input of heat from its surroundings to proceed. Understanding enthalpy change is crucial because it directly affects the Gibbs free energy and helps predict the feasibility of reactions.
Entropy Change
Entropy, represented by \( \Delta S \), refers to the degree of disorder or randomness in a system. Entropy change measures the change in this disorder as a result of a chemical reaction. It is measured in joules per Kelvin per mole.A positive \( \Delta S \) indicates an increase in disorder, while a negative \( \Delta S \) suggests a decrease. In our specific reaction, \( \Delta S \) is \( 105 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1} \), a positive number. This means that the products (gaseous \( \mathrm{BrCl} \)) are more disordered than the reactants (liquid \( \mathrm{Br}_{2} \) and gaseous \( \mathrm{Cl}_{2} \)).The change in entropy plays a vital role in determining the spontaneity of a process. A reaction with a significant increase in entropy can proceed spontaneously even if it absorbs heat (endothermic), as seen in processes like the melting of ice.
Reaction Equilibrium
Chemical equilibrium occurs when a reaction's forward and backward rates are equal, resulting in no net change in the concentration of reactants and products over time. At equilibrium, the Gibbs free energy change, \( \Delta G \), is zero.In the context of the Gibbs equation, \( \Delta G = \Delta H - T \Delta S \), equilibrium is reached when the enthalpy change of the system balances out the product of the temperature and the entropy change. Thus, no further reaction occurs as the system's energy is minimized.Understanding the concept of equilibrium is key to predicting and controlling chemical processes. It not only determines the extent of reactions but also influences conditions such as pressure and concentration that favor either forward or reverse reactions.
Temperature Calculation
Calculating the temperature at which a chemical reaction reaches equilibrium involves using the Gibbs free energy equation rearranged as \( T = \frac{\Delta H}{\Delta S} \). To find this temperature, use the given values of enthalpy and entropy change.In the problem, the enthalpy change \( \Delta H \) is first converted from kilojoules to joules \((30,000 \, \text{J} \, \text{mol}^{-1})\).Then, dividing \( \Delta H \) by \( \Delta S \), \[ T = \frac{30,000 \, \text{J} \, \text{mol}^{-1}}{105 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}} \approx 285.71 \, \text{K} \]This calculation shows the specific temperature at which the system will reach equilibrium. At this point, the effect of the enthalpy and entropy changes are balanced, and there is no net energy change, making it a critical parameter for understanding reaction conditions.

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Most popular questions from this chapter

The latent heat of vaporization of a liquid at \(500 \mathrm{~K}\) and 1 atm pressure is \(10.0 \mathrm{kcal} / \mathrm{mole}\). The change in internal energy of one mole of the liquid at the same temperature and pressure is _________ kcal.

The enthalpies of solution of \(\mathrm{BaCl}_{2}\) (s) and \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (s) are \(-20.6\) and \(8.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The enthalpy change for the hydration of \(\mathrm{BaCl}_{2}(\mathrm{~s})\) is (a) \(29.8 \mathrm{~kJ}\) (b) \(-11.8 \mathrm{~kJ}\) (c) \(-20.6 \mathrm{~kJ}\) (d) \(-29.4 \mathrm{~kJ}\).

For which of the following reactions, is \(\Delta \mathrm{H}\) equal to \(\Delta \mathrm{E} ?\) (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HI}(\mathrm{g})\) (b) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (c) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) (d) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\)

At \(300 \mathrm{~K}\) and \(1 \mathrm{~atm}, 15 \mathrm{~mL}\) a gaseous hydrocarbon requires \(375 \mathrm{~mL}\) air containing \(20 \% \mathrm{O}_{2}\) by volume for complete combustion. After comustion the gases occupy \(330 \mathrm{~mL}\). Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: (a) \(\mathrm{C}_{3} \mathrm{H}_{8}\) (b) \(\mathrm{C}_{4} \mathrm{H}_{8}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{10}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{6}\)

Identify the state function among the following: (a) \(\mathrm{Q}\) (b) \(\mathrm{Q}-\mathrm{w}\) (c) \(\mathrm{Q} / \mathrm{w}\) (d) \(\mathrm{Q}+\mathrm{w}\)

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