Chemistry help please? Calculate the enthalpy change for the reaction: C6H4(OH)2(aq) + H2O2(aq) <—>C6H4O2(aq) + 2H2O(l)?

Question

Calculate the enthalpy change for the reaction: C6H4(OH)2(aq) + H2O2(aq) <—> C6H4O2(aq) + 2H2O(l)

given the following data:

C6H4(OH)2(aq)-->C6H4O2(aq)+H2(g) H= +174.4 kJ

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

H2(g)+1/2O2(g)-->H2O(g) H= -241 kJ



H2(g)+O2(g)-->H2O2(aq) H= -191.2 kJ 

Answer 1

HESS' LAW:

C6H4(OH)2(aq) + H2O2(aq) <—> C6H4O2(aq) + 2H2O(l) ... TARGET EQUATION

eq.1. C6H4(OH)2(aq) --> C6H4O2(aq)+H2(g) H = +174.4 kJ

eq.2. H2O(g) --> H2O(l) H = -43.8 kJ

eq.3. H2(g)+1/2O2(g) --> H2O(g)   H = -241 kJ

eq.4. H2(g)+O2(g) --> H2O2(aq)   H = -191.2 kJ  

Procedure:

copy eq.1: C6H4(OH)2(aq) --> C6H4O2(aq)+H2(g) H = +174.4 kJ

reverse eq.4: H2O2(aq) --> H2(g) + O2(g) H = +191.2 kJ

copy eq.3x2: 2H2(g)+O2(g) --> 2H2O(g)   H = -482 kJ

copy eq.2x2: 2H2O(g) --> 2H2O(l) H = -87.6 kJ

Add them all up and combine or cancel like items:

C6H4(OH)2(aq) + H2O2(aq) ==> C6H4O2(aq) + 2H2O(l) ... TARGET EQUATION

∆H = 174.4 + 191.2 - 482 - 87.6 = -204 kJ

Answer 2

Hello Tzuyu

Flip the last equation( when you flip the equation the heat change sign must be changed to positive)

H2O2(aq)--> H2(g)+O2(g) H= +191.2 kJ 

Add the first equation

C6H4(OH)2(aq)-->C6H4O2(aq)+H2(g) H= +174.4 kJ

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C6H4(OH)2(aq) + H2O2(aq) --> C6H4O2(aq)+ 2H2(g) + O2(g) H= +365.6 kj

Multiply the two middle equations with two and add all the equations

2H2O(g)-->2H2O(l) H= -87.6 kJ

2H2(g)+ O2(g)-->2H2O(g) H= -482 kJ

Adding all the equations

C6H4(OH)2(aq) + H2O2(aq) + 2H2O(g) + 2H2(g)+ O2(g) -->

2H2O(g) + 2H2O(l) + C6H4O2(aq)+ 2H2(g) + O2(g) H= -204

C6H4(OH)2(aq) + H2O2(aq) --> 2H2O(l) + C6H4O2(aq) H= -204