ThermochemistryMedium
Question
The standard enthalpies of formation of NH3(g), CuO(s) and H2O(l) are −46, −155 and −285 kJ/mol, respectively. The enthalpy change when 6.80 g of NH3 is passed over cupric oxide is
Options
A.−59.6 kJ
B.+59.6 kJ
C.−298 kJ
D.−119.2 kJ
Solution
$2NH_{3}(g) + 3CuO(s) \rightarrow 3Cu(s) + N_{2}(g) + 3H_{2}O(l)$
$\Delta_{r}H = \sum_{}^{}{\Delta_{f}H_{\text{Products}} - \sum_{}^{}{\Delta_{f}H_{\text{Reactants}}}} $$${= \left\lbrack 0 + 0 + 3 \times ( - 285) \right\rbrack - \left\lbrack 2 \times ( - 46) + 3 \times ( - 155) \right\rbrack = - 298\text{ kJ} }{\therefore\Delta H_{\text{required}} = - \frac{298}{2 \times 17} \times 6.8 = 59.6\text{ kJ}}$$
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