A) \[x=y\]
B) \[x=\frac{y}{2}\]
C) \[~x=2y\]
D) None of these
Correct Answer: B
Solution :
Heat of neutralisation = Heat of formation of water (1 mol) \[HCl+NaOH\xrightarrow{{}}NaCl+{{H}_{2}}O;\] \[\Delta H=-x\] or \[\underset{1\,mol}{\mathop{{{H}^{+}}}}\,+\underset{1\,mol}{\mathop{O{{H}^{-}}}}\,\xrightarrow{{}}\underset{1\,mol}{\mathop{{{H}_{2}}O;}}\,\] \[\Delta H=x\] \[{{H}_{2}}S{{O}_{4}}+2NaOH\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O;\Delta H=-y\]or \[2{{H}^{+}}+2O{{H}^{-}}\xrightarrow{{}}2{{H}_{2}}O;\] \[\Delta H=-y\] \[{{H}^{+}}+O{{H}^{-}}\xrightarrow{{}}{{H}_{2}}O;\] \[\Delta H=-\frac{y}{2}\] \[\Rightarrow \] \[\Delta H=\Delta H\] \[\therefore \] \[x=\frac{y}{2}\]
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