C6H4(OH)2 (aq) + H2O2 (aq) C6H402 (aq) + 2 H2O (l) GIVEN THE REACTIONS BELOW, CALCULATE THE ∆H OF NET REACTION OF C6H4(OH)2 (aq) + H2O2 (aq) C6H402 (aq) + 2 H2O (l) GIVEN THE FOLLOWING REACTIONS GET ∆H FOR THE ABOVE REACTIONS OF THE MECHANISM. C6H4(OH)2 (aq) C6H4O2(aq) + H2(g) ∆H = +177.4 kJ H2(g) + O2(g) H2O2(aq) ∆H = -191.2 kJ H2(g) + ½ O2 H2O(g) ∆H = -241.8 kJ H2O(g)  H2O(l) ∆H = -43.8 kJ 1) Reverse the reactions such that reactants are on the left and products are on the right. 2) Cross cancel or cross reduce coefficients. 3) You can only cross cancel the same phase of the same species. 4) You must get the net reaction from the mechanism steps.

C6H4(OH)2 (aq) C6H4O2(aq) + H2(g) ∆H = +177.4 kJ H2(g) + O2(g) H2O2(aq) ∆H = -191.2 kJ H2(g) + ½ O2 H2O(g) ∆H = -241.8 kJ H2O(g)  H2O(l) ∆H = -43.8 kJ Must reverse this reaction to get the peroxide as a reactant and reverse the sign of the ∆H

C6H4(OH)2 (aq) C6H4O2(aq) + H2(g) ∆H = +177.4 kJ H2O2(aq)  H2(g) + O2(g) ∆H = =+191.2 kJ 2H2(g) + O2(g)  2H2O(g) ∆H =2(-241.8 kJ) 2H2O(g)  2H2O(l) ∆H =2(-43.8 kJ) Doubled this reaction and its ∆H, this will allow the oxygen to cancel with the oxygen in reaction 4