The heat of neutralization of oxalic acid is $- 53.35 k J e q u i v^{- 1}$ using NaOH. Hence, $Δ H$ of
H₂C₂O₄ ⇌ C₂O₄²⁻+ 2H⁺ is

Question

The heat of neutralization of oxalic acid is − 53.35 k J e q u i v − 1 using NaOH. Hence, Δ H of H₂C₂O₄ ⇌ C₂O₄²⁻+ 2H⁺ is

TestHub · Accepted Answer

By the definition of heat of neutralization, we have:

$\frac{1}{2} H_{2} C_{2} O_{4} + N a O H ⟶ \frac{1}{2} N a_{2} C_{2} O_{4} + H_{2} O$ ; $Δ H = - 53.35 k J$

or

$\frac{1}{2} H_{2} C_{2} O_{4} + O H^{-} ⟶ \frac{1}{2} C_{2} O_{4}^{2 -} + H_{2} O$ ; $Δ H = - 53.35 k J$ …(i)

$H^{+} + O H^{-} ⟶ H_{2} O$ ; $Δ H = - 57.3 k J$ …(ii)

Subtracting equation (ii) from (i), we get:

$\frac{1}{2} H_{2} C_{2} O_{4} ⟶ \frac{1}{2} C_{2} O_{4}^{2 -} + H^{+}$ ; $Δ H = 3.95 k J$

Therefore,

$H_{2} C_{2} O_{4} ⟶ C_{2} O_{4}^{2 -} + 2 H^{+}$ ; $Δ H = 7.9 k J m o l^{- 1}$

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