An exam question requires students to calculate delta Hdegree f for CIF3(g) give

Question

An exam question requires students to calculate delta Hdegree f for CIF3(g) given the reaction: and that delta Hdegree f HF(g) = -271.0 kJ/mol and deltaHdegreef NH3(g) = -46.00 kJ/mol While attempting this problem, a student incorrectly reads the table of delta H degree rxn values provided and uses -1212 kJ for delta H degree rxn. He then correctly performs the algebra required to solve this problem, but gets the question wrong due to the fact that he used the incorrect delta H degree rxn in his calculations. What value of deltaH degree f CIF3(g) did the student calculate? Calculate your answer to the correct number of significant figures. kJ/mol

Explanation / Answer

the given reaction is

2ClF3 + 2NH3 ---> N2 + 6HF + Cl2

we know that

dHo rxn = dHfo products - dHfo reactants

so

dHo rxn = dHfo N2 + ( 6 x dHfo HF ) + dHfo Cl2 - ( 2 x dHfo ClF3) - ( 2 x dHfo NH3)

given

dHfo HF = -271

dHfo NH3 = -46

dHfo N2 = 0

dHfo Cl2 = 0

dHo rxn = -1212

so

using those values

we get

dHo rxn = dHfo N2 + ( 6 x dHfo HF ) + dHfo Cl2 - ( 2 x dHfo ClF3) - ( 2 x dHfo NH3)

-1212 = 0 + ( 6 x -271 ) + 0 - ( 2 x dHfo ClF3) - ( 2 x -46)

so

2 x dHfo ClF3 = -322

dHfo clF3 = -161 kJ/mol

so

dHfo ClF3 value calculated by the student is -161 kJ/mol

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