ThermochemistryHard
Question
The thermochemical equation for the dissociation of hydrogen gas into atoms may be written as
H2 → 2H; ΔH = 432.0 kJ.
What is the ratio of the energy yield on combustion of hydrogen atoms to steam to the yield on combustion of an equal mass of hydrogen molecules to steam? Heat of formation of steam is −240.0 kJ/mol.
Options
A.2.80
B.1.80
C.0.8
D.2.40
Solution
$H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(g);\Delta H_{1} = - 240\text{ kJ}$
$2H(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(g);\Delta H_{2} = - (240 + 432) = - 672\text{ kJ}$
$\therefore\frac{\Delta H_{2}}{\Delta H_{1}} = \frac{- 672}{- 240} = 2.8$
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