Question

Using standard heats of formation, calculate the standard enthalpy change for the following reaction 2HBr(g) H2(g) + Br2() ANSWER kJ Submit Answer Retry Entire Group 2 more group attempts remaining

Answer 1

1)

Given:

Hof(HBr(g)) = -36.4 KJ/mol

Hof(H2(g)) = 0.0 KJ/mol

Hof(Br2(l)) = 0.0 KJ/mol

Balanced chemical equation is:

2 HBr(g) ---> H2(g) + Br2(l)

ΔHo rxn = 1*Hof(H2(g)) + 1*Hof(Br2(l)) - 2*Hof( HBr(g))

ΔHo rxn = 1*(0.0) + 1*(0.0) - 2*(-36.4)

ΔHo rxn = 72.8 KJ

Answer: 72.8 KJ

2)

Lets number the reaction as 1, 2, 3 from top to bottom

required reaction should be written in terms of other reaction

This is Hess Law

required reaction can be written as:

reaction 3 = +1 * (reaction 1) -1 * (reaction 2)

So, ΔHo rxn for required reaction will be:

ΔHo rxn = +1 * ΔHo rxn(reaction 1) -1 * ΔHo rxn(reaction 2)

= +1 * (-415.0) -1 * (-341.8)

= -73.2 KJ

Answer: -73.2 KJ

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