Question

If heat of dissociation of CHCl2COOH is 0.7 kcal/mole then $\Delta$H for the reaction :

CHCl2COOH + KOH $\longrightarrow$CHCl2COOK + H2O

• A. – 13 kcal
• B.
  • 13 kcal
• C. – 14.4 kcal
• D. – 13.7 kcal

Answer 1

Answer: (A)

– 13 kcal

Answer 2

H+ + OH– $\longrightarrow$ H2O ($\mathcal{l}$)

$\Delta$H = – 13.7 kcal ...(i)

CHCl2COOH $\longrightarrow$ CHCl2COO– + H+

$\Delta$H = 0.7 kcal ...(ii)

CHCl2COOH + OH– $\longrightarrow$

CHCl2COO– + H2O($\mathcal{l}$)$\Delta$H = ?

Adding equation (i) and (ii) $\Delta$H = – 13 kcal.