Can someone explain in detail all of the steps to do thisproblem? Given the aver

Question

Can someone explain in detail all of the steps to do thisproblem? Given the average bond dissociation enthalpies, estimate theclosest value to the standard molar enthalpy of formation ofClF3 (gas). Answer: -405 kJ/mol F-F 155kJ/mol Cl-F 253 kJ/mol Cl-Cl 242 kJ/mol Can someone explain in detail all of the steps to do thisproblem? Given the average bond dissociation enthalpies, estimate theclosest value to the standard molar enthalpy of formation ofClF3 (gas). Answer: -405 kJ/mol F-F 155kJ/mol Cl-F 253 kJ/mol Cl-Cl 242 kJ/mol

Explanation / Answer

First, balance your equation. Cl2 + F2 --------> ClF3 Balanced:    Cl2 + 3F2--------> 2ClF3 Second, use the formula that total molar enthalpy equals the sum ofbonds broken minus the sum of bonds formed. H = Bonds broken - Bonds formed Next, using your balanced equation, find all of the bonds that arebroken and all of the bonds that are formed. Broken: one Cl-Cl bond, 3 F-F bonds Formed: 6 Cl-F bonds Plug these values and coefficients into the equation: H = [(1*242kj/mol) + (3*155kj/mol)] - (6*253kj/mol) H = (242kj/mol + 465kj/mol) - 1518kj/mol H = -811kJ/mol Given the balanced equation, we have solved for the enthalpy fortwo mols, but we need it for one. -811kJ/mol /2 = -405.5 kJ/mol

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