You visited us 1 times! Enjoying our articles? Unlock Full Access!

Other Synthesis Methods

When 1 mole o...

Question

When 1 mole of oxalic acid is treated with excess of NaOH in dilute aqueous silution, 108 kJ of heat is liberated, then the enthalpy of ionisation of the oxalic acid is (Given, $H^{+}$ (aq)+ $O H^{-}$ (aq) $\to$ $H_{2} O$ (I); $△$ _neut H = -57.3 kJ)

4.6 kJ mol

^{-}

^{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-4.6 kJ mol

^{-}

^{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-6.6 kJ mol

^{-}

^{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.6 kJ mol

^{-}

^{1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 6.6 kJ mol $^{-}$ $^{1}$
$C_{2} H_{2} O_{4} + 2 N a O H (a q .) \to N a_{2} C_{2} O_{4} (a q .) + 2 H_{2} O (l)$
ionic reaction / equation ahoms are
$C_{2} H_{2} O_{4} (a q .) + 2 O H (a q .) \to C_{2} H_{4}^{2 -} (a q .) + 2 H_{2} O (a q .) . . . .$ equation $(1)$
enthalpy for above reaction of $H^{+}$ is written as $H^{+} (a q .) +^{-} O H (a q .) \to H_{2} O (l) - - - - (2)$
newtralisition enthalpy of above $H^{4} = 57.3 K I$
Let resonance equation $(2)$ and multiply it by $2$
new equation we get
$2 H_{2} O (l) + 2 H^{+} (a q) + 2^{-} O H (a q .) - - - (3)$
enthalpy chance for above $r c^{n} = 2 \times 57.9 = 114.6 K I$ (reduced $r n$ are)
add equatio $(1)$ & $(2)$
$C_{2} H_{2} O_{4} (a q .) \to 2 H^{+} (a q .) + C_{2} O_{4}^{2 -} (a q .) - - - Δ H = 6.6 K I$
Therefore enthalpy of conjugation of oxalic acid is $6.6 K I$

Suggest Corrections

Similar questions

When 1 mole of oxalic acid is treated with excess of NaOH in dilute aqueous solution, 106 kJ of heat is liberated. Predict the enthalpy of ionization of the acid ?

Q. Internal energy change for combustion of oxalic acid (

H_{2} C_{2} O_{4}

) is

- 245.7 k J / m o l

. The enthalpy of the combustion of oxalic acid at

27^{o}

C is:

Q. Consider the following reactions.

(a)

H_{(a q)}^{+} + O H_{(a q)}^{-} = H_{2} O_{(l)}, Δ H = - X_{1}

m o l^{- 1}

(b)

H_{2 (g)} + \frac{1}{2} O_{2 (g)} = H_{2} O_{(l)}, Δ H = X_{2}

m o l^{- 1}

(c)

C O_{2 (g)} + H_{2 (g)} = C O_{(g)} + H_{2} O_{(l)} - X_{3}

m o l^{- 1}

(d)

C_{2} H_{2 (g)} + \frac{5}{2} O_{2 (g)} = 2 C O_{2 (g)} + H_{2} O_{(l)} + X_{4}

m o l^{- 1}

Enthalpy of formation of

H_{2} O_{(l)}

is:

Q. Enthalpy of neutralisation of acetic acid by

N a O H

- 50.6 k J m o l^{- 1}

. Calculate

Δ H

for ionisation of

C H_{3} C O O H

. Given, the heat of neutralisation of a strong acid with a strong base is

- 55.9 k J m o l^{- 1}

Join BYJU'S Learning Program

When 1 mole of oxalic acid is treated with excess of NaOH in dilute aqueous silution, 108 kJ of heat is liberated, then the enthalpy of ionisation of the oxalic acid is (Given, H+(aq)+ OH−(aq) → H2O(I); △_neut H = -57.3 kJ)

When 1 mole of oxalic acid is treated with excess of NaOH in dilute aqueous silution, 108 kJ of heat is liberated, then the enthalpy of ionisation of the oxalic acid is (Given, $H^{+}$ (aq)+ $O H^{-}$ (aq) $\to$ $H_{2} O$ (I); $△$ _neut H = -57.3 kJ)