Question

Calculate ΔHrxn for the following reaction: 
CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g) 
Use the following reactions and given ΔH′s. 
C(s)+2H2(g)→CH4(g)     ΔH=−74.6kJ
C(s)+2Cl2(g)→CCl4(g)    ΔH=−95.7kJ
H2(g)+Cl2(g)→2HCl(g)   ΔH=−184.6kJ

Answer 1

CH₄(g)+4Cl₂(g)?CCl₄(g)+4HCl(g) (reaction 4)

Use the following reactions and given ?H?s.
C(s)+2H₂(g)?CH₄(g) (reaction 1) ?H₁=?74.6kJ
C(s)+2Cl₂(g)?CCl₄(g) (reaction 2) ?H₂=?95.7kJ
H₂(g)+Cl₂(g)?2HCl(g) (reaction 3) ?H₃=?184.6kJ

To find reaction 4 from the combination of the other 3 reactions (Hess's Law):

CH₄(g) ? C(s)+2H₂(g) (reverse reaction 1) - ?H₁=+74.6 kJ

C(s)+2Cl₂(g)?CCl₄(g) (reaction 2) ?H₂=?95.7 kJ

2H₂(g)+2Cl₂(g)?4HCl(g) (2 x reaction 3) 2x?H₃=?369.2kJ

If you add and delete what exist in right and left you will find:

CH₄(g)+4Cl₂(g)?CCl₄(g)+4HCl(g) (reaction 4 = (- reaction 1) + (reaction 2) + (2xreaction 3))

$\Rightarrow$ ?H₄ = - ?H₁ + ?H₂ +2 x ?H₃ = +74.6 + (-95.7) + (-369.2) = - 390.3 kJ