Consider the generic reaction: $$ \mathrm{A}+2 \mathrm{~B} \longrightarrow \mathrm{C} \Delta H_{\mathrm{ran}}=-55 \mathrm{~kJ} $$ 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) \(1 \mathrm{~mol} \mathrm{~A}\) (b) 2 mol \(\mathrm{A}\) (c) \(1 \mathrm{~mol} \mathrm{~B}\) (d) \(2 \mathrm{~mol} \mathrm{~B}\)

Short Answer

Expert verified
The amount of heat emitted for (a) 1 mol A is -55 kJ, (b) 2 mol A is -110 kJ, (c) 1 mol B is -27.5 kJ, (d) 2 mol B is -55 kJ.

Step by step solution

01

Identify the Heat of Reaction for the Given Equation

Firstly, it's important to understand that the heat of reaction, ewline ewline ewline or enthalpy change (ewline ewline ewline H), provided (ewline ewline ewline H_{ran} = -55 kJ), corresponds to the amounts of reactants specified by the coefficients in the balanced chemical equation. Here, for every mole of A reacting with 2 moles of B, 55 kJ of heat is released.
02

Calculate the Heat Emitted for 1 mole of A

Since the given reaction releases -55 kJ per mole of A reacting with 2 moles of B, and part (a) presents a scenario with 1 mole of A, we apply the stoichiometry directly to find the heat emitted: ewline ewline ewline Q = -55 kJ
03

Calculate the Heat Emitted for 2 moles of A

For part (b), with 2 moles of A reacting, simply multiply the heat emitted for 1 mole of A by 2: ewline ewline ewline Q = 2 ewline ewline ewline (-55 kJ) = -110 kJ
04

Calculate the Heat Emitted for 1 mole of B

For part (c), we need to adjust for the stoichiometry of B. Since it takes 2 moles of B to release the same -55 kJ of heat, for 1 mole of B, the heat released would be half that amount: ewline ewline ewline Q = ewline ewline ewline H_{ran}/2 = (-55 kJ)/2 = -27.5 kJ
05

Calculate the Heat Emitted for 2 moles of B

Finally, for part (d), since 2 moles of B is the stoichiometric amount needed to react with 1 mole of A to release -55 kJ of heat, the result is the same for 1 mole of A: ewline ewline ewline Q = -55 kJ

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.

Chemical Reactions
Chemical reactions are the processes by which substances, known as reactants, transform into different substances called products. Understanding the nature of chemical reactions is essential for predicting how compounds will interact under various conditions.

In a reaction, the alignment of atoms and molecules changes, which can result in a release or absorption of energy. The reaction in the given exercise is an example of this principle, where reactant A combines with reactant B to form a new product C. In such reactions, the law of conservation of mass tells us that no atoms are lost or gained; they are simply rearranged to create new compounds.

Identifying Reactants and Products

When looking at a chemical equation, it's crucial to identify the reactants and products first. In our exercise, A and B are the reactants, whereas C is the product. The coefficients in front of the reactants, such as the '2' in front of B, indicate the stoichiometry, or the ratio of reactants that react together.
Stoichiometry
Stoichiometry is the branch of chemistry that quantitatively relates the amounts of substances consumed and produced in chemical reactions. It is based on the law of conservation of mass and the concept of the mole.

With stoichiometry, we can calculate the amounts of reactants needed to produce a desired quantity of product, or vice versa. For instance, the provided exercise showcases a balanced chemical reaction in which 1 mole of A reacts with 2 moles of B to produce C.

Importance of Balance

The equation must be balanced to apply stoichiometry correctly, meaning the number of atoms of each element must be the same on both the reactant and product sides. Our exercise features a balanced equation, which is critical for accurate calculations. Students should always ensure equations are balanced before applying stoichiometric principles.
Heat of Reaction
The heat of reaction, or the enthalpy change (H), is the amount of heat absorbed or released during a chemical reaction at constant pressure. It's an essential aspect of thermochemistry, as it helps predict energy changes during reactions.

In the given problem, the H is indicated as -55 kJ, representing an exothermic reaction where heat is released into the surroundings when the reactants A and B are converted to product C. The negative sign signifies the release of heat.

Exothermic vs. Endothermic

Exothermic reactions release heat, as our example does, while endothermic reactions absorb heat. This distinction significantly affects how chemical processes are managed in various industrial applications and natural phenomena.
Thermochemistry
Thermochemistry studies the relationships between chemical reactions and energy changes involving heat. It's concerned with the measurement and understanding of the energy changes that accompany reactions.

The concepts of enthalpy, entropy, and Gibbs free energy are central to this field. In practical terms, knowing the thermochemical aspects of a reaction can inform us about the feasibility of a process and its potential impact on the environment.

Applications

Thermochemistry is used in designing energy-efficient processes in the chemical industry, understanding metabolic pathways in biology, and in environmental impact assessments. It plays a key role in the development of new materials and energy sources, emphasizing the importance of calculating and interpreting heat changes in reactions.

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

If you are given a chemical equation and specific amounts for each reactant in grams, how do you determine the maximum amount of product that can be made?

Consider the reaction: $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ If a reaction mixture contains \(28 \mathrm{~g}\) of \(\mathrm{CO}\) and \(32 \mathrm{~g}\) of \(\mathrm{O}_{2}\), what is the limiting reactant? (Try to do this problem in your head without any written calculations.)

For each reaction, calculate how many moles of product form when \(1.75 \mathrm{~mol}\) of the reactant in color completely reacts. Assume there is more than enough of the other reactant. (a) \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g)\) (b) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)\) (c) \(2 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (d) \(2 \mathrm{~S}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\)

For the reaction shown, calculate how many moles of \(\mathrm{NO}_{2}\) form when each amount of reactant completely reacts. $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ (a) \(1.3 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\) (b) \(5.8 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\) (c) \(4.45 \times 10^{3} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\) (d) \(1.006 \times 10^{-3} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\)

Consider the generic chemical reaction: $$ \mathrm{A}+2 \mathrm{~B} \longrightarrow \mathrm{C} $$ How many moles of \(C\) are formed upon complete reaction of: (a) \(2 \mathrm{~mol}\) of \(\mathrm{A}\) (b) \(2 \mathrm{~mol}\) of \(\mathrm{B}\) (c) \(3 \mathrm{~mol}\) of \(\mathrm{A}\) (d) \(3 \mathrm{~mol}\) of \(\mathrm{B}\)

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