consider the folliwing again?

Question

Given the following data
C6H4(OH)2 (aq) --> C6H4O2 (aq) + H2 (g) delta H = 177.4 kJ

H2 (g) + O2 (g) --> H2O2 (aq) delta H = -191.2 kJ

H2 (g) + 1/2O2 (g) --> H2O (g) delta H = -241.8 kJ

H2O (g) --> H2O (l) delta H = -43.8 kJ

Calculate the enthalpy change (in kJ) for the reaction

C6H4(OH)2 (aq) + H2O2 (aq)--> C6H4O2 (aq) + 2H2O (l)

Answer 1

I haven't worked this type of problem for a long problem, but I think you can do it as follows:

If you number the 4 "given" equations 1 through 4, then the final equation consists of:
1 example of equation 1
1 example of equation 2 in reverse (delta H = +191.2 kJ)
2 examples of equation 3 (using the H2 and O2 produced in the 2 preceding equations)
2 examples of equation 4 (using the H2O produced in the preceding equation)

So you should be able to get the answer by calculating:
[1] - [2] + 2 [3] + 2 [4]
(where [1] represents the delta H for equation 1)

Good luck!