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  • Enthalpy of neutralisation of H3PO3 is using NaOH. If ent

Enthalpy of neutralisation of H3PO3 is \mathrm{-110 kJ/mol} using NaOH. If enthalpy of neutralisation of HCl by NaOH is \mathrm{-57.3 kJ/mol}. Find the change in enthalpy of ionisation of H3PO3 into its ions( in \mathrm{kJ/mol}).

Option: 1

57.3


Option: 2

-57.3


Option: 3

4.6


Option: 4

-4.6


Answers (1)

best_answer

Writing the reactions:


\mathrm{HCl(aq) + NaOH (aq) \rightarrow H_2O + NaCl \;\;\;\; \Delta H_1 = -57.3\ kJ}

\mathrm{H_3PO_3(aq) + 2NaOH (aq) \rightarrow Na_2HPO_3 +2H_2O(l) \;\;\;\; \Delta H_2 = -110kJ}

We have to find the enthalpy of ionization of Phosphorous acid, which means we have to find the enthalpy of reaction :

\mathrm{H_3PO_3(aq) \rightarrow HPO_3^{-2} +2H^{+} \;\;\;\; \Delta H_3 = ?}

For solving this question, let's eliminate the spectator ions, namely Cl and Na ions.
Now the reactions will be written as:
 

\mathrm{H^+ + OH^-\rightarrow H_2O(l) \;\;\;\; \Delta H_1 = - 57.3\ kJ}    \mathrm{H_3PO_3 (aq) + 2OH^- \rightarrow HPO_3^{-2} + 2H_2O(l) \;\;\;\; \Delta H_2 = -110kJ/mol}

Now multiply equation (i) by 2 and subtract this from equation (ii) to get the desired reaction.
 

\mathrm{\Delta H_3 = \Delta H _2 - 2\times \Delta H_1}

\mathrm{\Delta H_3 = -110 -2\times (-57.3) }

\mathrm{\Rightarrow \Delta H_3 = 4.6\ kJ\ mol^{-1}}

Posted by

Rakesh

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