At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO 3 : 2 KClO 3 (s) → 2 KCl(s) + 3 O 2 (g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (a) 1.36 mol of O 2

hey everyone in this example, we have the following reaction And we need to find the entropy of formation of 5.15 g of lead chloride. So we want to recognize whether our equation is balanced and based on our psyche aama tree and the coefficients. It's definitely a balanced equation. Our next step is to recognize that we are given the entropy of formation for reaction for our formation of lead chloride as negative 3 36 point okay, joules per mole and this is for one mole of our lead chloride. So we want to recognize that that comes from the coefficient one which is in front of our product lead chloride. So our next step is to first because we are given the amount in grams for lead chloride. We need to convert from grams to moles. So we want to start from 5.15 g of our lead chloride. And then we're going to go from grams to moles by recalling our molar mass for the compound. So when we find lead in group four a on our periodic tables, we see that it has a molar mass of 207.2 g per mole. And when we find chlorine in group seven a on the periodic table, We see that it has a molar mass of 35.4, 5 g per mole. Now recognize that in our formula, we have two moles of our Chlorine Atom. And so we're going to multiply this by two. So to get the total mass for the compound lead chloride, we're going to have these two masses added up together And we're going to have a total of 278.1 g per mole for our compound P BCL two. So we're going to plug that in. So we have to 78.1 g for one mole of our lead chloride. So this allows us to cancel out our units of grams leaving us with moles, which is what we want. And we're going to get a value equal to 0.01815 moles of our lead chloride. And actually I re I just need to rewrite this number. So it's actually 0.018 519 moles of our lead chloride. And so now we can calculate our anthem P A formation for our lead chloride For our 5.15 g of lead chloride. And we would get that that is equal to our entropy of formation for one mole, which is given as negative 3 36.0 Killer Jewels for one mole of lead chloride. And then we're multiplying this by our molds of our 5.15 g of our lead chloride which we found above. So we plugged that in as 0. moles of our lead chloride. So this allows us to cancel out most, leaving us with kayla jewels as our final unit, Which is definitely what we need for entropy of formation. And we get a value in our calculators equal to negative 6.22238. So we can round this to about 366 so that we have negative 6.22 kg joules as our final answer For the entropy of formation of our 5. g of our product, lead chloride. So I hope that everything we reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Ch.5 - Thermochemistry

All textbooks

Brown 14th Edition

Ch.5 - Thermochemistry

Problem 46a

Chapter 5, Problem 46a

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (a) 1.36 mol of O₂

Verified step by step guidance

Identify the given reaction: 2 KClO_3(s) → 2 KCl(s) + 3 O_2(g) with ΔH = -89.4 kJ.

Understand that ΔH = -89.4 kJ is for the formation of 3 moles of O_2.

Set up a proportion to find ΔH for 1.36 moles of O_2: (ΔH for 1.36 mol O_2) / (ΔH for 3 mol O_2) = 1.36 mol / 3 mol.

Solve the proportion to find ΔH for 1.36 moles of O_2.

Ensure the units are consistent and the sign of ΔH reflects the exothermic nature of the reaction.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate the amounts of substances consumed and produced in a reaction based on balanced chemical equations. In this case, the stoichiometric coefficients from the reaction indicate that 3 moles of O2 are produced for every 2 moles of KClO3 decomposed.

Enthalpy Change (ΔH)

Enthalpy change (ΔH) is a measure of the heat content of a system at constant pressure. It indicates whether a reaction is exothermic (releases heat, ΔH < 0) or endothermic (absorbs heat, ΔH > 0). In the given reaction, ΔH = -89.4 kJ signifies that the decomposition of KClO3 releases 89.4 kJ of energy for every 2 moles of KClO3 reacted, which is crucial for calculating the energy change for the formation of 1.36 moles of O2.

Molar Relationships

Molar relationships refer to the ratios of moles of reactants and products in a balanced chemical equation. These relationships are essential for converting between moles of different substances. In this scenario, knowing that 3 moles of O2 are produced from 2 moles of KClO3 allows us to determine the amount of energy associated with the formation of 1.36 moles of O2 by using the established stoichiometric ratios.

Recommended video:

Guided course

01:13

Molarity

Related Practice

Textbook Question

Consider the following reaction: 2 CH₃OH(g) → 2 CH₄(g) + O₂(g) ΔH = +252.8 kJ (c) For a given sample of CH₃OH, the enthalpy change during the reaction is 82.1 kJ. How many grams of methane gas are produced?

Textbook Question

Consider the following reaction: 2 CH₃OH(g) → 2 CH₄(g) + O₂(g) ΔH = +252.8 kJ (d) How many kilojoules of heat are released when 38.5 g of CH₄(g) reacts completely with O₂(g) to form CH₃OH(g) at constant pressure?

458

views

Textbook Question

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates Ag⁺(aq) + Cl^-(aq) → AgCl(s) H = -65.5 kJ (a) Calculate H for the production of 0.450 mol of AgCl by this reaction. (b) Calculate H for the production of 9.00 g of AgCl. (c) Calculate H when 9.25⨉10^-4 mol of AgCl dissolves in water.

861

views

Textbook Question

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (b) 10.4 g of KCl.

423

views

Textbook Question

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ (c) The decomposition of KClO₃ proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of KClO₃ from KCl and O₂, is likely to be feasible under ordinary conditions? Explain your answer.

559

views

Textbook Question

Consider the combustion of liquid methanol, CH₃OH(l): CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l) ΔH = -726.5 kJ (a) What is the enthalpy change for the reverse reaction?

743

views

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3: 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (a) 1.36 mol of O2

Verified video answer for a similar problem:

Key Concepts

Stoichiometry

Enthalpy Change (ΔH)

Molar Relationships

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (a) 1.36 mol of O₂