Given the following thermochemical reaction: 2 B(s) + 3 H(g) ---> B2H6(g) dH = +

Question

Given the following thermochemical reaction: 2 B(s) + 3 H(g) ---> B2H6(g) dH = +36 kJ How many moles of boron (B) would be needed in this reaction if the dH is +18 kJ? Select one: A. 2 mol B. 4 mol C. 1mol True of False: The reverse reaction of 2 B(s) + 3 H(g) ---> B2H6(g) dH = +36 kJ is an exothermic reaction. Select one: True False Using the following thermochemical equation, answer the following questions. N2(g) + 2O2(g) --> 2 NO2(g) dH = +68 kJ If the reaction amounts were increased by a factor of 3, how would the delta H be affected? What would be the delta H of the reverse reation? How much heat is generated when 4 mol of NO2 is produced?

Explanation / Answer

a) Moles of boron needed for +18 J dH = 2*(18/36)

                                                      = 1 mole Boron

b) For reverse reaction dH = -36KJ

thus it is exothermic reaction .

Ans = TRUE

c) i) dH would be 3 times the initial

    ii) For reverse reaction dH = -68 KJ

    iii) When 4 moles NO2 is produced ,

         Heat = 68*(4/2)= 68*2 = 136 kJ

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