/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Consider the following hypotheti... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following hypothetical reactions: $$ \begin{array}{ll} \mathrm{A} \rightarrow \mathrm{B} & \Delta H=+30 \mathrm{~kJ} \\ \mathrm{~B} \longrightarrow \mathrm{C} & \Delta H=+60 \mathrm{~kJ} \end{array} $$ (a) Use Hess's law to calculate the enthalpy change for the reaction \(\mathrm{A}-\cdots \mathrm{C}\) (b) Construct an enthalpy diagram for substances \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\), and show how Hess's law applies.

Short Answer

Expert verified
(a) Using Hess's law, we calculate the enthalpy change for the reaction A to C as follows: \[ \Delta H_\text{A to C} = \Delta H_\text{A to B} + \Delta H_\text{B to C} = (+30 \mathrm{~kJ}) + (+60 \mathrm{~kJ}) = +90 \mathrm{~kJ} \] (b) The enthalpy diagram visually demonstrates that the enthalpy change for the reaction A to C is equal to the sum of the enthalpy changes for the reactions A to B and B to C, in accordance with Hess's law.

Step by step solution

01

Review Hess's law

Hess's law states that the overall change in enthalpy for a given reaction is equal to the sum of the individual enthalpy changes for the steps of that reaction when added together. In other words, the enthalpy change of a reaction is path-independent and depends only on the initial and final states.
02

Calculate the enthalpy change for the reaction A to C using Hess's law

According to Hess's law, we can find the enthalpy change for the reaction A to C by adding the enthalpy changes for the two given reactions. So, the enthalpy change for the reaction A to C can be calculated as follows: \[ \Delta H_\text{A to C} = \Delta H_\text{A to B} + \Delta H_\text{B to C} \] Substitute the given enthalpy changes: \[ \Delta H_\text{A to C} = (+30 \mathrm{~kJ}) + (+60 \mathrm{~kJ}) \] \[ \Delta H_\text{A to C} = +90 \mathrm{~kJ} \]
03

Construct the enthalpy diagram

To construct an enthalpy diagram, plot the enthalpy changes on the vertical axis and the substances on the horizontal axis. Begin with substance A at an arbitrary starting enthalpy, then plot the increase in enthalpy as we go from A to B, and similarly for B to C.
04

Show how Hess's law applies to the enthalpy diagram

On the enthalpy diagram, we can see that the overall enthalpy change for the reaction A to C is equal to the sum of the enthalpy changes for the reactions A to B and B to C, which is in accordance with Hess's law. In summary, the enthalpy change for the reaction A to C is +90 kJ, and the constructed enthalpy diagram visually demonstrates how Hess's law applies to this problem.

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Key Concepts

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

Enthalpy
Enthalpy, often represented as \( \Delta H \), is a key concept in understanding energy changes during chemical reactions.
It refers to the heat content of a system and is an important factor in determining how energy is absorbed or released as reactions occur.Think of enthalpy as the total energy held within a chemical system.
This includes the internal energy of the molecules and the pressure-volume work the system might do as it expands or contracts.
When a chemical reaction occurs, the enthalpy change \( \Delta H \) tells us if the reaction releases energy (exothermic, \( \Delta H < 0 \)) or absorbs energy (endothermic, \( \Delta H > 0 \)).In the example provided, when substance A turns into substance B, the change in enthalpy is \(+30 \mathrm{~kJ}\).
And further, from B to C, the change is \(+60 \mathrm{~kJ}\).
These values illustrate how energy is absorbed in each step of the reaction, ultimately summing to a total change.
Understanding these changes helps us predict the energy profile of reactions under different conditions.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products through the breaking and forming of bonds.
These processes can occur in a single step or through multiple intermediates, sometimes employing catalysts to speed them up. In any given chemical reaction, the enthalpy changes because of the conversion of reactants (A to B, and B to C in our case).
This breakage and recombination of chemical bonds necessitate energy input or output, which enthalpy change denotes.
The hypothetical reactions show how energy is managed stepwise as substance A converts to substance B, and subsequently from B to C. Using chemical reactions to illustrate Hess's law, we see that regardless of whether it happens directly (A to C) or through an intermediary (A to B to C), the total enthalpy change remains unchanged.
This principle shows why chemical reactions can be pieced out into steps structurally, giving insight into complex processes like burning fuel or digesting food.
Path Independence
One of the most intriguing aspects of Hess's Law is the concept of path independence.
Path independence means that the overall change in enthalpy for a process is independent of how that process occurs.With the A to C reaction example, path independence is clear.
Whether the reaction proceeds in a single step or through an intermediate stage does not affect the final change in enthalpy, which sums to \(+90 \mathrm{~kJ}\).
It's as if we are saying that only the starting point (A) and ending point (C) matter, not the stations in-between.Path independence is crucial in thermodynamics as it validates safely breaking complex reactions into manageable steps and understanding energy transformations without having to investigate every intermediate stage.
Consequently, this principle of path independence under Hess's Law enables scientists and engineers to design efficient processes, simplifying the study of energy exchanges in laboratories and industries alike.

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Most popular questions from this chapter

The complete combustion of acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}(l)\), to form \(\mathrm{H}_{2} \mathrm{O}(l)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases \(871.7 \mathrm{~kJ}\) of heat per mole of \(\mathrm{CH}_{3} \mathrm{COOH}\). (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

Two solid objects, \(\mathrm{A}\) and \(\mathrm{B}\), are placed in boiling water and allowed to come to temperature there. Each is then lifted out and placed in separate beakers containing \(1000 \mathrm{~g}\) water at \(10.0^{\circ} \mathrm{C}\). Object \(\mathrm{A}\) increases the water temperature by \(3.50^{\circ} \mathrm{C} ; \mathrm{B}\) increases the water temperature by \(2.60{ }^{\circ} \mathrm{C}\). (a) Which object has the larger heat capacity? (b) What can you say about the specific heats of \(\mathrm{A}\) and \(\mathrm{B}\) ?

Calculate \(\Delta E\), and determine whether the process is endothermic or exothermic for the following cases: (a) A system absorbs \(105 \mathrm{~kJ}\) of heat from its surroundings while doing \(29 \mathrm{~kJ}\) of work on the surroundings; (b) \(q=1.50 \mathrm{~kJ}\) and \(w=-657 \mathrm{~J} ;\) (c) the system releases \(57.5 \mathrm{~kJ}\) of heat while doing \(22.5 \mathrm{~kJ}\) of work on the surroundings.

Thestandard enthalpies of formation of gaseous propyne \(\left(\mathrm{C}_{3} \mathrm{H}_{4}\right)\), propylene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)\), and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) are \(+185.4,+20.4\), and \(-103.8 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) Calculate the heat evolved per mole on combustion of each substance to yield \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) (b) Calculate the heat evolved on combustion of \(1 \mathrm{~kg}\) of each substance. (c) Which is the most efficient fuel in terms of heat evolved per unit mass?

Under constant-volume conditions the heat of combustion of benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is \(26.38 \mathrm{~kJ} / \mathrm{g}\). A \(1.640\) \(g\) sample of benzoic acid is burned in a bomb calorimeter. The temperature of the calorimeter increases from \(22.25^{\circ} \mathrm{C}\) to \(27.20^{\circ} \mathrm{C}\). (a) What is the total heat capacity of the calorimeter? (b) A 1.320-g sample of a new organic substance is combusted in the same calorimeter. The temperature of the calorimeter increases from \(22.14^{\circ} \mathrm{C}\) to \(26.82^{\circ} \mathrm{C}\). What is the heat of combustion per gram of the new substance? (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

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