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731c63f0-388c-4ddc-a8bc-30a23a39d0b8 Application on Hess’s Law medium Summary Question 1: Given the following reactions and their respective enthalpy changes: C(s) + O2(g) → CO2(g) ΔH1 = -393.5 kJ/mol H2(g) + 1/2O2(g) → H2O(l) ΔH2 = -286.0 kJ/mol C(s) + H2(g) → CH4(g) ΔH3 = -74.8 kJ/mol Calculate the enthalpy change for the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Answer 1: To calculate the enthalpy change for the given reaction, we can use Hess's Law. By manipulating the given reactions, we can cancel out the common compounds and add the enthalpy changes. Multiplying reaction 1 by 2 gives: 2C(s) + 2O2(g) → 2CO2(g) 2ΔH1 = 2(-393.5 kJ/mol) = -787.0 kJ/mol Multiplying reaction 2 by 2 gives: 2H2(g) + O2(g) → 2H2O(l) 2ΔH2 = 2(-286.0 kJ/mol) = -572.0 kJ/mol Adding reactions 3, 2, and 1 gives: C(s) + H2(g) + 2H2(g) + O2(g) + 2O2(g) → CH4(g) + 2H2O(l) + 2CO2(g) ΔH3 + 2ΔH2 + 2ΔH1 = -74.8 kJ/mol + (-572.0 kJ/mol) + (-787.0 kJ/mol) = -1433.8 kJ/mol Since the given reaction is the reverse of the calculated reaction, the enthalpy change for the given reaction is the negative of the calculated value. ΔH = -(-1433.8 kJ/mol) = 1433.8 kJ/mol Question 2: Given the following reactions and their respective enthalpy changes: 2SO2(g) + O2(g) → 2SO3(g) ΔH1 = -198.2 kJ/mol S(s) + O2(g) → SO2(g) ΔH2 = -296.8 kJ/mol 2S(s) + 3O2(g) → 2SO3(g) ΔH3 = -792.0 kJ/mol Calculate the enthalpy change for the reaction: 2SO2(g) + O2(g) → 2SO3(g) + 198.2 kJ Answer 2: To calculate the enthalpy change for the given reaction, we can use Hess's Law. By manipulating the given reactions, we can cancel out the common compounds and add the enthalpy changes. Multiplying reaction 2 by 2 gives: 2S(s) + 2O2(g) → 2SO2(g) 2ΔH2 = 2(-296.8 kJ/mol) = -593.6 kJ/mol Adding reactions 1 and 2 gives: 2SO2(g) + O2(g) + 2S(s) + 2O2(g) → 2SO3(g) + 2

b91d4948-db86-4074-b363-dd7a5f404e2c The reaction coordinate shows the progress of the reaction. Summary

Alkanes may seem difficult at first. The nomenclature, the structures, the reactions - it can all be overwhelming. But don't let that discourage you. Because every great organic chemist started where you are. They faced the same challenges and setbacks, but they never gave up. They never let the complexity defeat them. < Back Alkanes Alkanes may seem difficult at first. The nomenclature, the structures, the reactions - it can all be overwhelming. But don't let that discourage you. Because every great organic chemist started where you are. They faced the same challenges and setbacks, but they never gave up. They never let the complexity defeat them. Files Download DAY 12 Notes Alkanes The IGCSE Organic Chemistry Challenge 2 .pdf Download PDF • 6.14MB Notes Topic Exercise Answers DAY 13 .14 HW part 1 and 2 Alkanes The IGCSE Organic Chemistry Challenge .pdf Download PDF • 14.40MB answers DAY 13 .14 HW part 1 and 2 Alkanes The IGCSE Organic Chemistry Challenge .pdf Download PDF • 17.28MB Videos Previous Next

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Distinguishing properties solids liquids and gases,Describe structure solids liquids and gases particle separation arrangement motion type 1.Particulate Nature of Matter IGCSE CHEMISTRY CAMBRIDGE 1.1 States Of Matter Notes : State the distinguishing properties of solids liquids and gases Matter in all states is made up of particles Particles forming matter can be represented by small solid spheres. Back to Chapter 1 Menu Solid particles : have a regular arrangement and are close together They have a fixed pattern ; they only vibrate in their position. Liquid particles : have a random arrangement they are close together They have no fixed pattern ; they slide past each other Gas particles : have a random arrangement , they are far apart from each other and they have no fixed pattern or shape , they move rapidly and randomly. A Solid has a definite shape and volume but can not flow A Liquid has a definite volume but has no fixed shape as it takes the shape of the container and it can flow A Gas has no definite shape or volume , it can spread everywhere through the container List of Topics Download as PDF