Answer:
A. DH° = –36 kJ
Explanation:
It is possible to obtain DH° of a reaction by the sum of DH° of half reactions. The DH° of the reaction:
B₂H₆(g) → 2B(s) + 3H₂(g)
Could be obtained from:
<em>(1) </em>2B(s) + 1.5O₂(g) → B₂O₃(s) DH° = –1273kJ
<em>(2) </em>B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g) DH° = –2035kJ
<em>(3) </em>H₂(g) + 0.5O₂(g) → H₂O(g) DH° = –242kJ
The sum of (2) - (1) gives:
B₂H₆(g) + 1.5O₂(g) → 2B(s) + 3H₂O(g) DH° = -2035kJ - (-1273kJ) = -762kJ
Now, this reaction - 3×(3):
B₂H₆(g) → 2B(s) + 3H₂(g) DH° = -762kJ - (3×-242kJ) = -36kJ
Thus, right answer is:
<em>A. DH° = –36 kJ</em>
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Explanation:
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