Consider the reaction directly below and answer parts a and b. c6h4(oh)2 (l) + h2o2 (l) à c6h4o2 (l) + 2 h2o (l)
(a) use the following information to calculate dh° for the above reaction. show all work. dh° c6h4(oh)2 (l) à c6h4o2 (l) + h2 (g) +177.4 kj h2 (g) + o2 (g) à h2o2 (l) -187.8 kj h2 (g) + o2 (g) à h2o (l) -285.8 kj
(b) based on your answer to part a, above, would heat be gained or lost by the surroundings as the reaction at the very top of the page occurred at standard conditions?
(c) how many kilojoules of heat are given off when 20.0 g of h2o (l) are produced at standard conditions according to the reaction at the very top of the page? show all work.

a. -206,4kJ

b. Surroundings will gain heat.

c. -115kJ are given off.

Explanation:

It is possible to obtain ΔH of a reaction using Hess's law that consist in sum the different ΔH's of other reactions until obtain the reaction you need.

Using:

(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

(2) H₂(g) + O₂(g) → H₂O₂ (l) ΔH: -187.8 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH: -285.8 kJ

It is possible to obtain:

C₆H₄(OH)₂(l) + H₂O₂(l) → C₆H₄O₂(l) + 2H₂O(l)

From (1)-(2)+2×(3). That is:

(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

-(2) H₂O₂(l) → H₂(g) + O₂(g) ΔH: +187.8 kJ

2x(3) 2H₂(g) + O₂(g) → 2H₂O(l) ΔH: 2×-285.8 kJ

The ΔH you obtain is:

+177,4kJ + 187,8kJ - 2×285.8 kJ = -206,4kJ

b. When ΔH of a reaction is <0, the reaction is exothermic, that means that the reaction produce heat and the surroundings will gain this heat.

c. 20,0g of H₂O are:

20,0g× = 1,11 mol H₂O

As 2 moles of H₂O are produced when -206,4kJ are given off, when 1,11mol of H₂O are produced, there are given off:

1,11mol H₂O× = -115kJ

I hope it helps!

the answer is barium and copper bromide