Question

The enthalpy change $(\Delta H)$ for the neutralisation of $M\,HCl$ by caustic potash in dilute solution at $298\, K$ is :

• A. 68 kJ
• B. 65 kJ
• C. 57.3 kJ
• D. 50 kJ

Answer 1

Answer: (C)

57.3 kJ

Answer 2

The heat of neutralisation of strong acid by strong base is always constant. $(57.3 \,kJ)$ It is infact heat of formation of water by $H^{+}$ and $O H^{-}$ ions. $K ^{+} OH ^{-}+ H ^{+} Cl ^{-} \rightarrow K ^{+} Cl ^{-}+ H _{2} O$ or $OH ^{-}+ H ^{+} \rightarrow H _{2} O \Delta H =57.3\, kJ$ $\because KOH$ is strong base and $HCl$ is strong acid.