**Calculating the Average Molar Bond Enthalpy of a Carbon–Chlorine Bond in CCl₄**Given the following standard enthalpies of formation:- ΔHₒf [Cl(g)] = 121.3 kJ·mol⁻¹- ΔHₒf [C(g)] = 716.7 kJ·mol⁻¹- ΔHₒf [CCl₄(g)] = −95.7 kJ·mol⁻¹Calculate the average molar bond enthalpy of the carbon–chlorine bond in a CCl₄ molecule.\[ \Delta H_{\text{C–Cl}} = \boxed{\phantom{fill-in-the-blank}} \,\, \text{kJ·mol⁻¹} \]

Enthalpy of formation of a substance is defined as the heat change that takes place when one mole of a substance is formed from its element under a given condition of temperature and pressure. The value of enthalpy of formation of elements in a reaction is always considered as zero.Enthalpy of formation of 1 mole of carbon=716.7 KJ/mol Enthalpy of formation of 1 mole of chlorine=121.3 KJ/mol Enthalpy of formation of 4 moles of chlorine, =4×121.3 KJ/mol=485.2 KJ/mol Therefore, the total enthalpy of reactants, =716.7+485.2=1201.9 KJ/mol Enthalpy of formation of 1 mole of CCl4=-95.7 KJ/mol. The total bond enthalpy of product, =-95.7 KJ/mol∆Hrxn=1201.9-(-95.7)=1297.6KJ/mol The average bond enthalpy of the reaction, =∆Hrxn4=1297.6 KJ/mol4=324.4 KJ/mol

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Chemistry

Given that AH; [CI(g)] = 121.3 kJ-mol-1 AH¡ [C(g)] = 716.7 kJ-mol¬! AH; [CCL,(g)] = -95.7 kJ-mol-! calculate the average molar bond enthalpy of the carbon-chlorine bond in a CCI, molecule. AHc-cı = kJ-mol-!

Given that AH; [CI(g)] = 121.3 kJ-mol-1 AH¡ [C(g)] = 716.7 kJ-mol¬! AH; [CCL,(g)] = -95.7 kJ-mol-! calculate the average molar bond enthalpy of the carbon-chlorine bond in a CCI, molecule. AHc-cı = kJ-mol-!

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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