Question

Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)Δ?1=−241.8 kJ1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ...

Given these reactions, where X represents a generic metal or metalloid

1) H2(g)+12O2(g)⟶H2O(g)Δ?1=−241.8 kJ1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ

2) X(s)+2Cl2(g)⟶XCl4(s)Δ?2=+157.3 kJ2) X(s)+2Cl2(g)⟶XCl4(s)ΔH2=+157.3 kJ

3) 12H2(g)+12Cl2(g)⟶HCl(g)Δ?3=−92.3 kJ3) 12H2(g)+12Cl2(g)⟶HCl(g)ΔH3=−92.3 kJ

4) X(s)+O2(g)⟶XO2(s)Δ?4=−769.5 kJ4) X(s)+O2(g)⟶XO2(s)ΔH4=−769.5 kJ

5) H2O(g)⟶H2O(l)Δ?5=−44.0 kJ5) H2O(g)⟶H2O(l)ΔH5=−44.0 kJ

what is the enthalpy, Δ?,ΔH, for this reaction?

XCl4(s)+2H2O(l)⟶XO2(s)+4HCl(g)

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Answer #1

)H, Ca)02 (I) 0 (a) 4H 241.S KJ 2) X() 20 () -XClq () 4H2 = + 157,3 kJ 3)H2()C () HC (3) 92.3 kJ 4H4 769.5 kJ X 02 (3) Xo2 (Sxcl4 (5)2H20 () -- xo, (s) + GHCI () Tanel eactia 5 neac tions to ot the ta^et We wil neannage the дven lde isill do tti by mStep 5 To caucel the extra tenms to obtain the toget equahon odd it to the pnevious neaction (, multipty by 2 neverne equatio

> absolute lifesaver. nowhere in my notes did it tell me I had to cancel out the irrelevant equations. Thanks for the info on the final step!

Emma Parry Fri, Jan 14, 2022 10:54 PM

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Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)Δ?1=−241.8 kJ1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ...
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