The heat of neutralization of oxalic acid is -25.4 kcal mol -1 using strong base, NaOH. Hence, the enthalpy change of the process is H 2 C 2 O 4 leftharpoons 2 H ⊕+ C 2 O 4 2- is

Question

The heat of neutralization of oxalic acid is -25.4 kcal mol -1 using strong base, NaOH. Hence, the enthalpy change of the process is H 2 C 2 O 4 leftharpoons 2 H ⊕+ C 2 O 4 2- is (A) 2.0 kcal (B) -11.8 kcal (C) 1.0 kcal (D) -1.0 kcal

Tardigrade Admin · Accepted Answer

Oxalic acid has two ionisableH ⊕. Hence, expected heat of neutralization, if it behaves as a strong acid would have been =-13.7 × 2=-27.4 kcal mol -1 But experimental value =-25.4 kcal mol -1 ∴ Heat of ionization =2.0 kcal mol -1

Q. The heat of neutralization of oxalic acid is $- 25.4 k c a l m o l^{- 1}$ using strong base, $N a O H$ . Hence, the enthalpy change of the process is $H_{2} C_{2} O_{4} ⇌ 2 H^{\oplus} + C_{2} O_{4}^{2 -}$ is

$2.0 k c a l$

$- 11.8 k c a l$

$1.0 k c a l$

$- 1.0 k c a l$

Oxalic acid has two ionisableH $^{\oplus}$ . Hence, expected heat of neutralization, if it behaves as a strong acid would have been
$= - 13.7 \times 2 = - 27.4 k c a l m o l^{- 1}$
But experimental value $= - 25.4 k c a l m o l^{- 1}$
$∴$ Heat of ionization $= 2.0 k c a l m o l - 1$

Q. The heat of neutralization of oxalic acid is −25.4kcalmol−1 using strong base, NaOH. Hence, the enthalpy change of the process is H2​C2​O4​⇌2H⊕+C2​O4​2− is

2.0kcal

−11.8kcal

1.0kcal

−1.0kcal