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<li> Given:
<ol type="a">
<li>C<sub>graphite</sub> + O<sub>2</sub> → CO<sub>2</sub>({{serif|g}}) ( Δ''H'' = −393.5 kJ/mol) (direct step)</li>
<li>C<sub>graphite</sub> + 1/2 O<sub>2</sub> → CO({{serif|g}}) (Δ''H'' = −110.5 kJ/mol)</li>
<li>CO({{serif|g}}) +1/2 O<sub>2</sub> → CO<sub>2</sub>({{serif|g}}) (Δ''H'' = −283.0 kJ/mol)</li>
</ol>
Reaction (a) is the sum of reactions (b) and (c), for which the total Δ''H'' = −393.5 kJ/mol, which is equal to Δ''H'' in (a).
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<li>Given:
*B<sub>2</sub>O<sub>3</sub>({{serif|s}}) + 3H<sub>2</sub>O({{serif|g}}) → 3O<sub>2</sub>({{serif|g}}) + B<sub>2</sub>H<sub>6</sub>({{serif|g}}) (Δ''H'' = 2035 kJ/mol)
*H<sub>2</sub>O(
*H<sub>2</sub>({{serif|g}}) + 1/2 O<sub>2</sub>({{serif|g}}) → H<sub>2</sub>O(
*2B({{serif|s}}) + 3H<sub>2</sub>({{serif|g}}) → B<sub>2</sub>H<sub>6</sub>({{serif|g}}) (Δ''H'' = 36 kJ/mol)
Find the Δ<sub>f</sub>''H'' of:
*2B({{serif|s}}) + 3/2 O<sub>2</sub>({{serif|g}}) → B<sub>2</sub>O<sub>3</sub>({{serif|s}})
After multiplying the equations (and their enthalpy changes) by appropriate factors and reversing the direction when necessary, the result is:
*B<sub>2</sub>H<sub>6</sub>({{serif|g}}) + 3O<sub>2</sub>({{serif|g}}) → B<sub>2</sub>O<sub>3</sub>({{serif|s}}) + 3H<sub>2</sub>O({{serif|g}}) (Δ''H'' = 2035 × (−1) = −2035 kJ/mol)
*3H<sub>2</sub>O({{serif|g}}) → 3H<sub>2</sub>O(
*3H<sub>2</sub>O(
*2B({{serif|s}}) + 3H<sub>2</sub>({{serif|g}}) → B<sub>2</sub>H<sub>6</sub>({{serif|g}}) (Δ''H'' = 36 kJ/mol)
Adding these equations and canceling out the common terms on both sides, we obtain
*2B({{serif|s}}) + 3/2 O<sub>2</sub>({{serif|g}}) → B<sub>2</sub>O<sub>3</sub>({{serif|s}}) (Δ''H'' = −1273 kJ/mol)
</li>
</ol>
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