Question

Calculate the delta H rxn for the following reaction:
CH4(g)+4Cl2(g)-->CCl4(g)+4HCl

Use the following reactions and given delta H's:

1) C(s)+2H2(g)-->CH4(g)
delta H= -74.6 kJ

2) C(s)+2Cl2(g)-->CCl4(g)
delta H= -95.7 kJ

3) H2(g)+Cl2(g)-->2HCl(g)
delta H= -184.6 kJ

Express your answer using one decimal place.

Answer 1

We'll reverse the 1st reaction:
CH4 --> C + 2H2 ; ΔH = +74.6

We'll use the 2nd equation as is:
C + 2Cl2 --> CCl4 ; ΔH = -95.7

We'll multiply the 3rd equation by 2:
2H2 + 2Cl2 --> 4HCl ; ΔH = -369.2

Now if we add the chemical equations, we'll end up with
CH4(g)+4Cl2(g)-->CCl4(g)+4HCl
Then add all the ΔH's to solve for the ΔH of this reaction.

Hope this helps~ `u`