Chemistry - Thermochemistry Question with Solution | TestHub

Moles of NaOH = M $\times$ Volume in liters

$$0.2\times \frac{100}{1000}=0.02$$

Moles of HCl = $0.4\times \frac{100}{1000}=0.04$

Both HCl and NaOH are strong electrolytes.

$$\mathrm{NaOH}⟶{\mathrm{Na}}^{+}+{\mathrm{OH}}^{-}$$

$$0.02mol--$$

$$0.02mol0.02mol$$

$$\mathrm{HCl}⟶{\mathrm{H}}^{+}+{\mathrm{Cl}}^{-}$$

$$0.04mol--$$

$$0.04mol0.04mol$$

Now, a neutralization reaction takes place:

${\mathrm{H}}^{+}(\mathrm{aq})+{\mathrm{OH}}^{-}(\mathrm{aq})⟶{\mathrm{H}}_2\mathrm{O}(\ell )$

   0.04 mole    0.02 mole                                              

↑

(Limiting reagent)

When 1 mole OH^- neutralizes with H⁺ ⇒ 57.5 kJ energy is released

∴ 0.02 mole OH^- neutralizes with H⁺ ⇒ (57.5 × 0.02)kJ ⇒ 1.15 kJ

⇒ 1150 J energy is released.