what is the enthalpy, Delta H for this reaction?

Question

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

1) H2(g)+ 1/2 O2(g) —> H2O(g) Delta H₁= -241.8kJ

2) X(s)+2Cl2(g) —> XCl4(s) Delta H₂= 318.5kJ

3) 1/2 H2(g) + 1/2 Cl2(g) —> HCl(g) Delta H₃= -92.3kJ

4) X(s)+O2 (g) —> XO2(s) Delta H₄= -766.7kJ

5) H2O(g) —> H2O(l) Delta H₅= -44kJ

what is the enthalpy, Delta H for this reaction?

XCl4(s)+2H2O(l) —> XO2(s) + 4HCl(g)

Answer 1

Review Hess's (Hess') Law for more information on how to approach these problems

XCl₄(s) + 2H2O(l) —> XO₂(s) + 4HCl(g) TARGET EQUATION

To get XCl₄ on the left, reverse equation (2) and change sign of ∆H₂

XCl₄(s) ==> X(s) + 2Cl₂(g) ... ∆H = -318.5 kJ

To get XO2(s) on the right, copy equation (4)

X(s) + O2(g) ==> XO2(s) ... ∆H = -766.7 kJ

To get 4HCl on the right, copy equation 3 and multiply it by 4

2H2(g) + 2Cl2(g) ==> 4HCl(g) ... ∆H = -369.2 kJ

If we add all of this together, at this point, we have the following:

XCl4(s) + X(s) + O2(g) + 2H2(g) + 2Cl2(g) ==> X(s) + 2Cl2(g) + XO2(s) + 4HCl(g) which reduces to ...

XCl4(s) + O2(g) + 2H2(g) ==> XO2(s) + 4HCl(g)

To get rid of O2(g) and H2(g) on the left, we can reverse equation (1) and multiply by 2

2H2O(g) ==> 2H2(g) + O2(g) ... ∆H = +483.6 kJ

Add this to the above and we have...

XCl4(s) + 2H2O(g) ==> XO2(s) + 4HCl(g)

And finally, to convert the H2O(g) to H2O(l), reverse equation (5) and add it to the above

H2O(l) ==> H2O(g) ... ∆H = +44 kJ

XCl4(s) + 2H2O(g) + H2O(l) ==> XO2(s) + 4HCl(g) + H2O(g)

XCl4(s) + H2O(l) ==> XO2(s) + 4HCl = TARGET EQUATION

∆H = -318.5 kJ + -766.7 kJ + -369.2 kJ + 483.6 kJ + 44 kJ

∆H = -926.8 kJ