Question

From the following data at 25°C:
H2(g)+ Cl2(g) 2HCl(g) ΔH = -185 kJ
2H2(g)+ O2(g) 2H2O(g) ΔH = -483.7 kJ

a. Calculate H at 25°C for the reaction below:

4HCl(g)+ O2(g) 2Cl2 (g) + 2H2O (g)

Answer 1

any idea how to solve this ?

Answer 2

It's thermodynamics. Our professor gave this as an advance homework and we don't have a clue on how to solve this.

Answer 3

ok let's start .first we have to compare the equation whose H we have to calculate ..See in our final equation the number of moles of HCl and Chlorine is double as in eq 1 ..so first multiply whole eq 1 by 2 .

Answer 4

multiply delta H value of eq 1 too ...

Answer 5

So, I'll just multiply the first equation and it's delta H value by 2?

Answer 6

then u see that in our final equation HCl and O2 are on the left side and Cl2 is on right side so we have to subtract eq 1 from eq 2

2H2 + O2 ->2H2O delta H = -483.7 kJ .......eq 2
-2H2 - 2 Cl2 --> -4 HCl delta H = +370 ......eq 1 ( sign changed )
------------------------------------------------------------
4 HCl + O2 --> 2 Cl2 + 2H2O delta H = -113.7 kj

Answer 7

2H2 ( eq 2 ) cancelled with 2H2 of eq 1 ..

Answer 8

Can you explain why did the sign changed?

Answer 9

because we subtract

Answer 10

Oh. I get it. Thank you very much.

Answer 11

anytime :)