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Question: The enthalpy of formation of \(H_{2}O(l)\) is – 285.77 kJ mol<sup>–1</sup> and enthalpy of neutralis...

The enthalpy of formation of H2O(l)H_{2}O(l) is – 285.77 kJ mol–1 and enthalpy of neutralisation of strong acid and strong base is –56.07 kJ mol–1, what is the enthalpy of formation of OH– ion

A
  • 229.70kJ
B

–229.70kJ

C

+226.70kJ

D

–22.670kJ

Answer

–229.70kJ

Explanation

Solution

H+(aq)+OH−(aq)→H2O(l);H^{+}(aq) + OH^{-}(aq) \rightarrow H_{2}O(l); ΔH=−56.07kJ\Delta H = - 56.07kJ

ΔH=ΔHf(H2O)−[ΔHf(H+)+ΔHf(OH)−]\mathbf{\Delta}\mathbf{H =}\mathbf{\Delta}\mathbf{H}_{\mathbf{f}}\mathbf{(}\mathbf{H}_{\mathbf{2}}\mathbf{O)}\mathbf{-}\mathbf{\lbrack}\mathbf{\Delta}\mathbf{H}_{\mathbf{f}}\mathbf{(}\mathbf{H}^{\mathbf{+}}\mathbf{) +}\mathbf{\Delta}\mathbf{H}_{\mathbf{f}}\mathbf{(OH}\mathbf{)}^{\mathbf{-}}\mathbf{\rbrack}

[∵ΔHf(H+)=0]\mathbf{\lbrack}\mathbf{\because}\mathbf{\Delta}\mathbf{H}_{\mathbf{f}}\mathbf{(}\mathbf{H}^{\mathbf{+}}\mathbf{) = 0\rbrack}

−56.07=−285.77−(0+x)\mathbf{-}\mathbf{56.07 =}\mathbf{-}\mathbf{285.77}\mathbf{-}\mathbf{(0 + x)}

∴x\mathbf{\therefore x} =−285.77+56.07=−229.70kJ\mathbf{=}\mathbf{-}\mathbf{285.77 + 56.07 =}\mathbf{-}\mathbf{229.70kJ}