A) \[4.3\text{ }kJ\text{ }mo{{l}^{-1}}\]
B) \[-4.3\,kJ\,mo{{l}^{-1}}\]
C) \[8.6\text{ }kJ\,mo{{l}^{-1}}\]
D) \[-8.6\,kJ\,mo{{l}^{-1}}\]
Correct Answer: C
Solution :
[c] \[{{H}^{+}}(aq)+O{{H}^{-}}(aq)\rightleftharpoons {{H}_{2}}O(l);\] \[\Delta H=-57.3kJ\] .... (i) \[{{H}_{2}}{{C}_{2}}{{O}_{4}}+2O{{H}^{-}}\rightleftharpoons {{C}_{2}}O_{_{4}}^{-}+2{{H}_{2}}O;\Delta H=-106kJ\].... (ii) For the second reaction the value should have been \[2\times \left( -57.3 \right)=-114.6kJ\] The difference \[\left( 114.6-106 \right)=8.6\text{ }kJ\,mo{{l}^{-1}}\] is used to effect of the ionisation of oxalic acid.
You need to login to perform this action.
You will be redirected in
3 sec
