Question

Given the enthalpies of reaction below, determine the enthalpy change for the reaction: H2O (g) + C(s) --> CO (g) + H2 (g).
The enthalpies of reaction are:
H2 (g) + 1/2 O2 (g) --> H2O (g); deltaH = -242 kJ
2 CO (g) --> 2 C (s) + O2 (g); deltaH = +221.0 kJ
I have tried to solve this many times, but cannot seem to get the answer. I would appreciate steps to solve this.

Answer 1

You would get better responses in the Chemistry study group

Answer 2

Give me a sec and I will try to solve and explain

Answer 3

The answer is 131.5 kJ
I will try to make it simple so let me just state some things.
Z= H20
X=C
V=CO
B=H2
N=O2

You want this equation; Z+X -> V+B
Given B+ 1/2N -> Z =-242 kj 2V -> 2X + N = -221 kj

Flip the first equation so the Z is on the left side; Z -> B+ 1/2 N
Since you flipped it, you would make the value positive

Since you need V on the right side,flip the 2nd equation so it becomes 2X+N -> 2V and then divide both sids by 2 to make it X + 1/2N -> V

Now I will replace the varibles with their values

H20 -> H2+ [1/2]O2 =242 kj
[1/2]O2 +C -> CO = [-221/2] kj

As you see, the O2 will cancel out with each other and what is left is H20+C -> H2+CO

So now you just add the kj up!
242+[-221/2] = 110.5 kj

Answer 4

flip-> H2 (g) + 1/2 O2 (g) --> H2O (g) delta H = -242.0 kJ
flip-> 2 CO (g) --> 2 C (s) + )2 (g) delta H = +221.0 kJ

2 x [H2O (g) --> H2 (g) + 1/2 O2 (g) ] delta H = 2 * 242.0 = + 484 kJ
2 H2O (g) --> 2 H2 + O2 (g)

O2 (g) + 2 C (s) --> 2 CO (g) delta H = -221.0 kJ

Adding together: 2 H2O (g) --> 2 H2 (g) + O2 (g)
+ O2 (g) + 2 C (s) --> 2 CO (g)
(Cancel out O2 on the reactant and product side).
= 2 H2O (g) + 2 C (s) --> 2 H2 (g) + 2 CO (g)
divided by 2 & simplified = H2O (g) + C (s) --> H2 (g) + CO (g)

+484 kJ + (-221 kJ) = +263 /2 = 131.5

Answer 5

Did I do it right?

Answer 6

No. You would not want to multiply the first part. If you do the same steps again, as I did , then you would get the right answer.