Consider the generic reaction: $$ 2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow \mathrm{C} \Delta H_{\mathrm{rxn}}=-125 \mathrm{~kJ} $$ Determine the amount of heat emitted when each amount of reactant completely reacts (assume that there is more than enough of the other reactant). (a) \(2 \mathrm{~mol} \mathrm{~A}\) (b) \(3 \mathrm{~mol} \mathrm{~A}\) (c) \(3 \mathrm{~mol} \mathrm{~B}\) (d) \(5 \mathrm{~mol} \mathrm{~B}\)

Short Answer

Expert verified
(a) -125 kJ, (b) -187.5 kJ, (c) -125 kJ, (d) -208.33 kJ

Step by step solution

01

Identify the relationship between moles and heat of reaction

Given the reaction, every 2 moles of A that react release -125 kJ of heat, and every 3 moles of B that react also release -125 kJ of heat.
02

Calculate heat emitted for 2 moles of A

Since the given reaction already specifies that 2 moles of A release -125 kJ, no calculation is needed for part (a).
03

Calculate heat emitted for 3 moles of A

For every 2 moles of A, -125 kJ are released. We need to set up a proportion to determine the heat for 3 moles of A: \((2 \text{ moles } \text{A}): (-125 \text{ kJ}) = (3 \text{ moles } \text{A}): (x \text{ kJ})\). Solve for x to find the heat released for 3 moles of A. The equation becomes \(x = \frac{3 \times (-125 \text{ kJ})}{2}\).
04

Calculate heat emitted for 3 moles of B

Since the reaction specifies that every 3 moles of B release -125 kJ, no calculation is needed for part (c).
05

Calculate heat emitted for 5 moles of B

For every 3 moles of B, -125 kJ are released. We need to set up a proportion to determine the heat for 5 moles of B: \((3 \text{ moles } \text{B}): (-125 \text{ kJ}) = (5 \text{ moles } \text{B}): (x \text{ kJ})\). Solve for x to find the heat released for 5 moles of B. The equation becomes \(x = \frac{5 \times (-125 \text{ kJ})}{3}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Heat of Reaction
The term heat of reaction, denoted by \( \Delta H_{\mathrm{rxn}} \), is a fundamental concept in thermochemistry. It refers to the amount of heat energy that is either absorbed or released during a chemical reaction. In exothermic reactions, such as the one given in the exercise, heat is emitted and the heat of reaction has a negative value, indicating a release of energy.

Understanding the heat of reaction is crucial for predicting the energy changes in chemical processes. It's also essential in industries where controlling the temperature of reactions is important for safety and product stability. The exercise demonstrates that when 2 moles of A react with B, there is a release of -125 kJ of heat. This is an intrinsic property of the reaction and is independent of the reaction pathway.
Mole-to-Energy Calculations
To comprehend mole-to-energy calculations, it's imperative to grasp the concept of the mole, which is a unit in chemistry representing an amount of substance. When discussing the energy exchanged in a reaction, we often calculate how much energy is associated with the reaction of one mole of a substance.

This concept is used in the given exercise to convert between the number of moles of reactants A and B and the amount of heat released, as seen in the step by step solution. By setting up a proportion, we can predict the heat released for any given amount of reactants. For example, for 3 moles of A, we calculate the heat emitted based on the given ratio of moles of A to energy, allowing us to scale the reaction's heat emission to different amounts of reactants.
Chemical Reaction Stoichiometry
At the heart of predicting the outcomes of chemical reactions is chemical reaction stoichiometry. It involves the quantitative relationship between the reactants and products in a balanced chemical equation. Stoichiometry allows us to deduce the amount of reactants needed to form a certain amount of product, or vice versa.

In the context of our exercise, the stoichiometric coefficients (2 for A and 3 for B) inform us about the ratio in which the reactants combine to form the product C. \( \Delta H_{\mathrm{rxn}} \) is associated with these exact stoichiometric amounts. When performing calculations, correctly applying stoichiometry ensures precise understanding of the reaction's scale. Misinterpreting these ratios can lead to errors in calculating the heat of reaction or in predicting how much product will form.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the unbalanced equation for the neutralization of acetic acid: $$ \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q) $$ Balance the equation and determine how many moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to completely neutralize \(1.07 \mathrm{~mol}\) of \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\).

For the reaction shown, calculate the theoretical yield of the product in grams for each of the initial quantities of reactants. $$ \mathrm{Ti}(s)+2 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{TiF}_{4}(s) $$ (a) \(1.0 \mathrm{~g} \mathrm{Ti} ; 1.0 \mathrm{~g} \mathrm{~F}_{2}\) (b) \(4.8 \mathrm{~g} \mathrm{Ti} ; 3.2 \mathrm{~g} \mathrm{~F}_{2}\) (c) \(0.388 \mathrm{~g} \mathrm{Ti} ; 0.341 \mathrm{~g} \mathrm{~F}_{2}\)

A solution contains an unknown mass of dissolved silver ions. When potassium chloride is added to the solution, a white precipitate forms. The precipitate is filtered and dried and found to have a mass of \(212 \mathrm{mg}\). What mass of silver was in the original solution? (Assume that all of the silver was precipitated out of solution by the reaction.)

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants. $$ 4 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s) $$ (a) \(1 \mathrm{~mol} \mathrm{Cr} ; 1 \mathrm{~mol} \mathrm{O}_{2}\) (b) \(4 \mathrm{~mol} \mathrm{Cr} ; 2.5 \mathrm{~mol} \mathrm{O}_{2}\) (c) \(12 \mathrm{~mol} \mathrm{Cr} ; 10 \mathrm{~mol} \mathrm{O}_{2}\) (d) \(14.8 \mathrm{~mol} \mathrm{Cr} ; 10.3 \mathrm{~mol} \mathrm{O}_{2}\)

Consider the equation for the combustion of acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\), the main ingredient in nail polish remover: $$ \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) } \\ {\Delta H_{\text {nxn }}=-1790 \mathrm{~kJ}} \end{array} $$ If a bottle of nail polish remover contains \(155 \mathrm{~g}\) of acetone, how much heat is released by its complete combustion?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free