Given these reactions, where x represents a generic metal or metalloid 1) h2(g)+12o2(g)⟶h2o(g)δh1=−241.8 kj 1) h2(g)+12o2(g)⟶h2o(g)δh1=−241.8 kj 2) x(s)+2cl2(g)⟶xcl4(s)δh2=+356.9 kj 2) x(s)+2cl2(g)⟶xcl4(s)δh2=+356.9 kj 3) 12h2(g)+12cl2(g)⟶hcl(g)δh3=−92.3 kj 3) 12h2(g)+12cl2(g)⟶hcl(g)δh3=−92.3 kj 4) x(s)+o2(g)⟶xo2(s)δh4=−639.1 kj 4) x(s)+o2(g)⟶xo2(s)δh4=−639.1 kj 5) h2o(g)⟶h2o(l)δh5=−44.0 kj 5) h2o(g)⟶h2o(l)δh5=−44.0 kj what is the enthalpy, δh,δh, for this reaction? xcl4(s)+2h2o(l)⟶xo2(s)+4hcl(g)

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!

answer: they produce more slow than unicellular organisms.