Calculate \(\Delta H^{\circ}\) for the process \(\mathrm{Sb}(s)+\frac{5}{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SbCl}_{5}(s)\) from the following information: \(\mathrm{Sb}(s)+\frac{3}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SbCl}_{3}(s) \quad \Delta H^{\circ}=-314 \mathrm{kJ}\) \(\mathrm{SbCl}_{3}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SbCl}_{5}(s) \quad \Delta H^{\circ}=-80 \mathrm{kJ}\)

Short Answer

Expert verified
\( \Delta H^{\circ}_{\text{total}} = -394 \mathrm{kJ} \)

Step by step solution

01

Write down the enthalpy changes for the given reactions

First, recognize the enthalpy changes given for the reactions: For the process \(\mathrm{Sb}(s) + \frac{3}{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SbCl}_{3}(s)\), the enthalpy change is \(-314 \mathrm{kJ}\). For the reaction \(\mathrm{SbCl}_{3}(s) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SbCl}_{5}(s)\), the enthalpy change is \(-80 \mathrm{kJ}\).
02

Combine the reactions to form the desired process

To find the enthalpy change for the desired process \(\mathrm{Sb}(s) + \frac{5}{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SbCl}_{5}(s)\), combine the two given reactions by adding them together. Since the second reaction forms \(\mathrm{SbCl}_{5}(s)\) from \(\mathrm{SbCl}_{3}(s)\) and an additional mole of \(\mathrm{Cl}_{2}(g)\), simply add it to the first reaction.
03

Calculate the overall enthalpy change

Add the enthalpy changes of the individual reactions to find the total enthalpy change for the desired process: \(-314 \mathrm{kJ}\) for the first reaction and \(-80 \mathrm{kJ}\) for the second reaction. \[\Delta H^{\circ}_{\text{total}} = \Delta H^{\circ}_{1} + \Delta H^{\circ}_{2} = (-314 \mathrm{kJ}) + (-80 \mathrm{kJ})\]

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Key Concepts

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

Thermochemistry
Thermochemistry is the study of the heat energy associated with chemical reactions and physical transformations. It's a fundamental concept within thermodynamics that fisuses on how energy is transferred in the form of heat, particularly during chemical processes. When substances react chemically, they either absorb or release energy, often in the form of heat.

Understanding the heat involved in these reactions is crucial for predicting reaction behavior and for applications in various fields, including the design of energy-efficient processes in industries. This branch of chemistry uses principles such as conservation of energy and the quantitative relationship between heat and work.
Hess's Law
Hess's Law, named after Russian chemist Germain Hess, states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps the reaction is carried out in. This law is a direct consequence of enthalpy being a state function.

Hess's Law in Practice:

When you cannot measure the enthalpy change of a reaction directly, you can use Hess's Law to calculate it indirectly. This is done by adding the enthalpy changes of individual steps that lead from reactants to products. It's akin to taking different routes to reach the same destination. The total distance traveled (enthalpy change) will remain constant irrespective of the path taken.
Enthalpy Change
Enthalpy change, denoted as \(\Delta H\), is the measure of heat change during a process at constant pressure. It's an extensive property, meaning it depends on the substance's mass. The sign of \(\Delta H\) indicates whether the system absorbed heat (positive \(\Delta H\)) from the surroundings or released heat (negative \(\Delta H\)) to the surroundings.

Determining Enthalpy Change:

To figure out the enthalpy change for a reaction, one can either measure it experimentally through calorimetry or calculate it using Hess's Law and standard enthalpies of formation. It's an essential factor in determining the energy efficiency of chemical processes and the favorability of reactions.
Chemical Reactions
Chemical reactions involve the transformation of one set of chemical substances to another through the breaking and forming of chemical bonds. They are described by chemical equations that show the reactants transforming into products. Several factors such as temperature, pressure, and concentration affect how reactions proceed.

The study of chemical reactions encompasses understanding reaction mechanisms, rates, and equilibrium. Furthermore, energy changes that occur during reactions are a focal point in chemical thermodynamics, helping to elucidate why certain reactions happen and others don’t. Mastering the concepts behind chemical reactions is pivotal for predicting reaction outcomes and manipulation in practical applications.

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Most popular questions from this chapter

Both graphite and diamond burn. \(\mathrm{C}(s, \text { diamond })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) For the conversion of graphite to diamond: \(\mathbf{C}(s, \text { graphite }) \longrightarrow \mathbf{C}(s, \text { diamond })\) \(\Delta H^{\circ}=1.90 \mathrm{kJ}\) Which produces more heat, the combustion of graphite or the combustion of diamond?

Which compound in each of the following pairs has the larger lattice energy? Note: \(\mathrm{Mg}^{2+}\) and \(\mathrm{Li}^{+}\) have similar radii; O \(^{2-}\) and \(\mathrm{F}^{-}\) have similar radii. Explain your choices. (a) MgO or MgSe (b) LiF or MgO (c) \(\mathrm{Li}_{2} \mathrm{O}\) or \(\mathrm{LiCl}\) (d) Li_se or MgO

Calculate the standard molar enthalpy of formation of \(\mathrm{NO}(g)\) from the following data: \(\mathrm{N}_{2}(g)+2 \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=66.4 \mathrm{kJ}\) \(2 \mathrm{NO}(g)+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=-114.1 \mathrm{kJ}\)

A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories (16 kcal). What is the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g?

A 45-g aluminum spoon (specific heat 0.88 J/g ^ C) at 24 ^ C is placed in 180 mL (180 g) of coffee at 85 ^ C and the temperature of the two become equal. (a) What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water. (b) The first time a student solved this problem she got an answer of \(88^{\circ} \mathrm{C}\). Explain why this is clearly an incorrect answer.

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