Question

Given that H2(g) + F2(g) - > 2HF(g) delta H rxn = -546.6 kJ 2H2(g)+ O2(g) - > 2H2O(l) delta H rxn = 571.6 kJ

Answer 1

Answer:

ΔHrxn = -521.6 kJ

Explanation:

To do this, let's write the equations by separate:

H₂ + F₂ -------> 2HF        ΔH = -546.6 kJ

2H₂ + O₂ -------> 2H₂O   ΔH = -571.6 kJ

For these reactions, we want to get the following reaction:

2F₂ + 2H₂O -------> 4HF + O₂     ΔHrxn = ?

To do this, all we have to do is take the first two reaction and put them in the way to obtain the third reaction. When we look the final reaction we can see that the water is on the reactants, when originally it was on the product, while Florine is doubled. So all we have to do is rewrite the first two reactions, duplicate the first reaction, and reverse the second reaction, and that way we will get the final reaction:

1) (H₂ + F₂ -------> 2HF ) x2      ΔH = -546.6 kJ x 2

2) (2H₂O --------> 2H₂ + O₂)   ΔH = -571.6 kJ x -1

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1) 2H₂ +2F₂ -------> 4HF      ΔH = -1093.2 kJ

2) 2H₂O --------> 2H₂ + O₂   ΔH = +571.6 kJ      

Now we sum 1) and 2). In this way, hydrogen cancels out and we do the same with the enthalpy:

1) 2H₂ +2F₂ -------> 4HF      ΔH = -1093.2 kJ

2) 2H₂O --------> 2H₂ + O₂   ΔH = +571.6 kJ  

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2F₂ + 2H₂O -------> 4HF + O₂     ΔHrxn = -1093.2 + 571.6 = -521.6 kJ

So the enthalpy of this reaction is

ΔHrxn = -521.6 kJ