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

see full answer

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

$- 130 k J / m o l$

$- 416 k J / m o l$

$+ 54 kJ / mol$

$+ 156 k J / m o l$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

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

$= - 416 kJ$

Watch 3-min video & get full concept clarity