Help with enthalpy
 

1. The bombardier beetle uses the following explosive reaction as a defence mechanism. The product acts as a repellent.

C6H4(OH)2 + H202 -> C6H4O2 + 2 H20

Calculate the enthalpy change for the above reaction using the following information equations.
1. C6H4O2 + H2 -> C6H4(OH)2 delta Hr = -177.4 kJ
2. 2 H20 + H2 -> O2 + 2 H20 delta Hr = -189.0 kJ
3. 1/2 O2 + H2 -> H20 delta Hr = -285.8kJ


I understand that we have to come out with the net equation, but somehow my calculations aren't able to cross components out.

So you think you might be able to help me, and if so show me the work?
Thank you.[/img]

required: C6H4(OH)2 + H202 -> C6H4O2 + 2 H20

available:
1. C6H4O2 + H2 -> C6H4(OH)2 delta Hr = -177.4 kJ
2. 2 H20 + H2 -> O2 + 2 H20 delta Hr = -189.0 kJ
3. 1/2 O2 + H2 -> H20 delta Hr = -285.8kJ


your equations are not balanced

2. should (probably) read
2H202 -> O2 + 2 H20

if you don't copy the equations down correctly then you have no chance!

Yes, I've realized. And thank you for the prompt reply.

I had already figured out my mistake. :D