Question

Consider the reaction C6H4(OH)2 (l) + H2O2 (l) ⎯→ C6H4O2 (l) + 2 H2O (l)

(a) Use the following information to calculate ΔH° for the above reaction. Show all work. ΔH° C6H4(OH)2 (l) ⎯⎯→ C6H4O2 (l) + H2 (g) +177.4 kJ H2 (g) + O2 (g) ⎯⎯→ H2O2 (l) -187.8 kJ H2 (g) + € 1 2 O2 (g) ⎯⎯→ 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.

Answer 1

C6H4(OH)2 (l) --> C6H4O2 (l) + H2 (g)     dhrxn = -177.4 Kj

H2O2 (l) ----> H2 (g) + O2 (g)           dhrxn = +187.8 kj

2H2(g) + O2(g) ----> 2H2O(l)            dhrxn = 2*-285.8 kj

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

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dhRXN = (-177.4+187.8+(2*-285.8))

      = -561.2 Kj

As DHrxn is - ve , it is exothermic

c) No of mol of water = 20/18 = 1.11 mol

2mol Water = 561 kj

1.11 mol water =   561*1.11/2 = -311.35 kj