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 ?

$4.3 k J m o l^{- 1}$

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

$- 4.3 k J m o l^{- 1}$

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

$- 8.6 k J m o l^{- 1}$

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

$8.6 k J m o l^{- 1}$

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

Open in App

Solution

The correct option is D
$8.6 k J m o l^{- 1}$

$H_{2} C_{2} O_{4 (a q)} \to 2 H_{a q}^{+} + C_{2} O_{4 (a q)}^{2 -}; Δ H = x k J \frac{[H_{(a q) +}^{+} O H_{(a q)}^{-} H_{2} O_{(r) . Δ H = - 57.3 k J}] \times 2}{H_{2} C_{2} O_{4} + 2 O H^{-} \to 2 H_{2} O_{(r)} + C_{2} O_{4}^{2 -}, Δ H = x - 114.6! k J} B u t x - 114.6 = - 106 (g i v e n) \Rightarrow x = 8.6 k J m o l^{- 1}$

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. What is the heat of solution of sodium chloride if hydration energy of

N a^{+}

- 389.4 {kJ mol}^{- 1}

, hydration energy of

C l^{-}

- 382.3 {kJ mol}^{- 1}

and lattice energy of

N a C l = + 776 {kJ mol}^{- 1}

1 mole of $H_{2} S O_{4}$ is mixed with excess of NaOH in a dilute solution. The enthalpy change is nearly equal to

What is the amount of heat liberated (in Kcal) when one mole of aqueous NaOH reacts with one mole of aqueous HCl solution?

1 mole of $H_{2} S O_{4}$ is mixed with excess of NaOH in a dilute solution. The enthalpy change is nearly equal to

Join BYJU'S Learning Program

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 ?

The correct option is D 8.6 kJ mol−1 H2C2O4(aq)→2H+aq+C2O2−4(aq);ΔH=x kJ[H+(aq)+OH−(aq)H2O(r).ΔH=−57.3kJ]×2H2C2O4+2OH−→2H2O(r)+C2O2−4,ΔH=x−114.6!kJBut x−114.6=−106(given)⇒x=8.6kJ mol−1