Question
When 1 mole of Na(s) is dissolved in large volume of water at 298 K and 1 bar, 184 kJ/mol heat is released. When 1 mole of Na2O(s) is dissolved in large volume of water at 298 K and 1 bar, 238 kJ/mol is released. If the enthalpy of formation of water is −286 kJ/mol, then the enthalpy of formation of sodium oxide is
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Solution
$(a)Na(s) + H_{2}O(l) \rightarrow NaOH(aq) + \frac{1}{2}H_{2}(g);\Delta H = - 184\text{ kJ}$
$(b)Na_{2}O(s) + H_{2}O(l) \rightarrow 2NaOH(aq);\Delta H = - 238\text{ kJ} $$${(c)H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l);\Delta H = - 286\text{ kJ} }{\text{Required T.C.E. is }2Na(s) + \frac{1}{2}O_{2}(g) \rightarrow Na_{2}O(s);\Delta H = ? }{\text{From 2}(a) - (b) + (c); }{ \Delta H = 2( - 184) - ( - 238) + ( - 286) = - 416\text{ kJ}}$$
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