Question

From the following enthalpies of reaction,
4 HCl (g) + O2 (g) → 2 H2O (l) + 2 Cl2 (g) ∆H = -202.4 kJ/mol 1/2 H2 (g) + ½ F2 (g) → HF (l) ∆H = -600.0 kJ/mol H2 (g) + ½ O2 (g) → H2O (l) ∆H = -285.8 kJ/mol
Calculate ∆Hrxn for 2 HCl (g) + F2 (g) → 2 HF (l) + Cl2 (g) Just input a number.

Answer 1

Answer:

ΔH= -1015.4

Explanation:

(1) 4HCl (g)+O₂(g)→2H₂O(l) +2Cl₂(g)       ΔH= -202.4 KJ/mol

(2) $\frac{1}{2}H_2(g)+\frac{1}{2} F_2 \rightarrow HF(l)$                     ΔH=-600.0 KJ/mol

(3) $H_2(g)+\frac{1}{2} O_2(g) \rightarrow H_2O (l)$                 ΔH= -285.8 KJ/mol

2HCl (g)+ F₂(g)→ 2HF(l) +Cl₂ (g)    

{Equation (1) ÷2}+{ Equation (2)× 2}- Equation (3)

$2HCl (g)+\frac{1}{2} O_2+H_2+F_2-H_2-\frac{1}{2} O_2\rightarrow H_2O+Cl_2+2HF -H_2O$

ΔH= (-202.4÷2)+(-600×2)-(-285.8) = -1015.4

⇒2HCl (g)+ F₂(g)→ 2HF(l) +Cl₂ (g)      ΔH= -1015.4

Answer 2

Considering the Hess's Law, the enthalpy change for the reaction is -1015.4 kJ.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

2 HCl (g) + F₂ (g) → 2 HF (l) + Cl₂ (g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 4 HCl (g) + O₂ (g) → 2 H₂O (l) + 2 Cl₂ (g)     ∆H = -202.4 kJ/mol

Equation 2: ½  H₂ (g) + ½ F₂ (g) → HF (l)     ∆H = -600.0 kJ/mol

Equation 3: H₂ (g) + ½ O₂ (g) → H₂O (l) ∆H = -285.8 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 2 moles of HCl (g) on reactant side and it is present in first equation But you need to divide it by 2 to obtain the amount of moles that you need. And since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is divided by 2, the variation of enthalpy also.

Now, you need 2 moles of HF (l) on rpoduct side and it is present in second. equation But you need to multiply it by 2 to obtain the amount of moles that you need, because you have 1 mole of the compound. Since the equation is multiplied by 2, the variation of enthalpy also.

Finally, to cancel H₂ and H₂O, you need to reverse third equation (inverse it). When an equation is inverted, the sign of delta H also changes.  

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 HCl (g) + ½ O₂ (g) →  H₂O (l) + Cl₂ (g)     ∆H = -101.2 kJ

Equation 2: H₂ (g) + F₂ (g) → 2 HF (l)     ∆H = -1200.0 kJ

Equation 3:  H₂O (l) → H₂ (g) + ½ O₂ (g)     ∆H = 285.8 kJ

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

2 HCl (g) + F₂ (g) → 2 HF (l) + Cl₂ (g)     ΔH= -1015.4 kJ

Finally, the enthalpy change for the reaction is -1015.4 kJ.

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