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Magnetic Moment Examples

Heat of neutr...

Question

Heat of neutralisation of oxalic acid is $- 106.7 k J m o l^{- 1}$ using $N a O H$ hence $△ H$ of
$H_{2} C_{2} O_{4} \to C_{2} O_{4}^{2 -} + 2 H^{+}$ is

5.88 K J

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- 5.88 K J

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- 13.7 K c a l

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7.5 K J

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Solution

The correct option is A $5.88 K J$
Oxalic acid $= C_{2} H_{2} O_{4}$
$2 H^{+}$ neutralise $\to - 106.7 k g / m o l e$
for $H_{2} C_{2} O_{4} \to 2 H^{+} \to - 106.7 k J$
Convert to $K c a l$
$= - 106.7 \times 0.239 k c a l$
$\frac{1}{2} H_{2} C_{2} O_{4} + N a O H \to \frac{1}{2} N O_{2} C_{2} O_{4} + H_{2} O; △ H = - 53.35 k J$
or
$\frac{1}{2} H_{2} C_{2} O_{4} + O H^{-} \to \frac{1}{2} C_{2} O_{4}^{2 -} + H_{2} O △ H = - 53.35 k J . . . . (i)$
$H^{+} + O H^{-} \to H_{2} O △ H = 57.3 k J . . . . (i i)$
Subtracting eq (ii) from (i)
$\frac{1}{2} H_{2} C_{2} O_{4} \to \frac{1}{2} C_{2} O_{4}^{2 -} + H^{+}$
$△ H = - 3.95$
$H_{2} C_{2} O_{4} \to C_{2} O_{4}^{2 -} + 2 H^{+}$
$△ H = 2 \times 3.95 = 7.9 k J$ .

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