(a) The standard molar enthalpy of formation, Delta Hf_0, for diborane, B2H6 (g)
ID: 511441 • Letter: #
Question
(a) The standard molar enthalpy of formation, Delta Hf_0, for diborane, B2H6 (g), cannot be directly cause the compound cannot be prepared by the reaction of boron and hydrogen. It can be calculated, however, using the following reactions: 4B(s) + 3O2(g) 2 rightarrow B2O3(s) Delta H degree_rxn = - 2543.8 kJ 2H2(g) + O2(g) rightarrow 2H2O(g) Delta H degree_rxn Delta H degree _rxn = -484. kJ B2H6(g) + 3 O2(g) rightarrow B2O3(s) + 3H2O(g) Delta H degree_rxn = -2032.9 kJ Calculate Delta H degree f for B2H6(g).Explanation / Answer
The required reaction for formatio of B2H6 is
2B(s) + 3H2(g) -------------> B2H6(g)
This equation can be obtained from the given equations as
1/2 (equation1) +3/2(equation 2) - equation 3
Thus delta H f of required equation = 1/2 (-2543.8) +3/2(-484) -(-2032.9) kJ
= +35kJ
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