The enthalpy of solution, of sodium and sodium oxide in a large volume of water, are $- 184 kJ / mole$ and $- 238 k J / m o l$ , respectively. If the enthalpy of formation of water is $- 286 kJ / mol$ , then what is the enthalpy of formation of sodium oxide? All the e: thalpies are at $298 K$ and 1 bar pressure. [Given: Reaction involved are
$2 Na (s) + 2 H_{2} O (l) ⟶ 2 NaOH (a q) + H_{2} (g)$
${Na}_{2} O (s) + H_{2} O (l) ⟶ 2 NaOH (a q)$

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$- 130 k J / m o l$

$- 416 k J / m o l$

$+ 54 kJ / mol$

$+ 156 k J / m o l$

answer is C.

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Detailed Solution

Given thermochemical equations are :

$(a) Na (s) + H_{2} O (l) ⟶ NaOH (a q) + \frac{1}{2} H_{2} (g);$

$Δ H_{1} = - 184 kJ$

$(b) {Na}_{2} O (s) + H_{2} O (l) ⟶ 2 NaOH (a q);$

$Δ H_{2} = - 238 kJ$

$(C) H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l);$

$Δ H_{3} = - 286 kJ$

The required thermochemical equation is :

$2 Na (s) + \frac{1}{2} O_{2} (g) ⟶ {Na}_{2} O (s)$

From $2 \times (a) + (c) - (b),$

$Δ H = 2 \times (- 184) + (- 286) - (- 238)$

$= - 416 kJ$

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