Answer:
ΔH = -793,6 kJ
Explanation:
It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:
If half-reactions are:
1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ
2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ
3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ
4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ
5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ
The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:
(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ
+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ
-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ
-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ
-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ
= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)
Where ΔH is:
ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ
ΔH = -793,6 kJ
I hope it helps!
The enthalpy, ΔH, for the reaction XCl₄(s) + 2H₂O(l) → XO₂(s) + 4HCl(g) where X represent a generic metal or metalloid is -793.6 kJ. It was calculated from the enthalpies of the reactions (1) to (5).
The given reactions are:
1) H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH₁ = -241.8 kJ
2) X(s) + 2Cl₂(g) → XCl₄(s) ΔH₂ = +356.9 kJ
3) ¹/₂H₂(g) + ¹/₂Cl₂(g) → HCl(g) ΔH₃ = -92.3 kJ
4) X(s) + O₂(g) → XO₂(s) ΔH₄ = -639.1 kJ
5) H₂O(g) → H₂O(l) ΔH₅ = -44.0 kJ
We need to find the enthalpy for the reaction:
XCl₄(s) + 2H₂O(l) → XO₂(s) + 4HCl(g) ΔH =? (6)
We will need to do some algebraic operations to reactions (1) to (5) to get the desired reaction (6).
For the XCl₄(s) and H₂O(l) to be on the reactants side (reaction 6), we need to reverse reactions (2) and (5):
2) XCl₄(s) → X(s) + 2Cl₂(g) ΔH₂ = -(+356.9 kJ) = -356.9 kJ
5) H₂O(l) → H₂O(g) ΔH₅ = -(-44.0 kJ) = +44.0 kJ
Now, we need to reverse reaction (1):
1) H₂O(g) → H₂(g) + ¹/₂O₂(g) ΔH₁ = -(-241.8 kJ) = +241.8 kJ
And then, multiply reactions (1) and (5) by a factor of two and reaction (3) by a factor of four:
1) 2H₂O(g) → 2H₂(g) + O₂(g) ΔH₁ = +483.6 kJ
3) 2H₂(g) + 2Cl₂(g) → 4HCl(g) ΔH₃ = -369.2 kJ
5) 2H₂O(l) → 2H₂O(g) ΔH₅ = -(-44.0 kJ) = +88.0 kJ
Finally, by adding reactions (1) to (5) we can eliminate the compounds H₂O(g), H₂(g), O₂(g), X(s), and Cl₂(g):
1) 2H₂O(g) → 2H₂(g) + O₂(g) ΔH₁ = +483.6 kJ
2) XCl₄(s) → X(s) + 2Cl₂(g) ΔH₂ = -(+356.9 kJ) = -356.9 kJ
3) 2H₂(g) + 2Cl₂(g) → 4HCl(g) ΔH₃ = -369.2 kJ
4) X(s) + O₂(g) → XO₂(s) ΔH₄ = -639.1 kJ
5) 2H₂O(l) → 2H₂O(g) ΔH₅ = -(-44.0 kJ) = +88.0 kJ
2H₂O(g) + XCl₄(s) + 2H₂(g) + 2Cl₂(g) + X(s) + O₂(g) + 2H₂O(l) → 2H₂(g) + O₂(g) + X(s) + 2Cl₂(g) + 4HCl(g) + XO₂(s) + 2H₂O(g)
By canceling all the repeated compounds in the reactants and products side, we have:
XCl₄(s) + 2H₂O(l) → XO₂(s) + 4HCl(g) (6)
The enthalpy of the above reaction is:
[tex] \Delta H = +483.6 kJ + (-356.9 kJ) + (-369.2 kJ) + (-639.1 kJ) + 88.0 kJ = -793.6 kJ [/tex]
Therefore, the enthalpy for reaction (6) is -793.6 kJ.
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I hope it helps you!
