Question

a) Calculate heat of dissociation for Acetic acid from the following data:

$C{H}_{3}COOH+NaOH⟶C{H}_{3}COONa+H_{2}O....\Delta H=-13.2Kcal$

$H^{\oplus }+\stackrel{\ominus }{O}H⟶H_{2}O;....\Delta H=-13.7Kcal$.

b) calculate heat of dissociation for $N{H}_{4}OH$ if

$HCl+N{H}_{4}OH⟶N{H}_{4}Cl+H_{2}O;\Delta H=-12.27Kcal$.

• A. a) $0.5kcal$ , b)$1.43cal$
• B. a) $-0.5kcal$ , b)$-1.43cal$
• C. a) $0.0kcal$ , b)$2.86cal$
• D. None of these

Answer 1

Answer: (B)

a) $-0.5kcal$ , b)$-1.43cal$

$\Delta H=\Delta H_{products}-{\Delta H}_{reactants}$

There is no enthalpy during the formation of salt.

$C{H}_{3}COOH+NaOH⟶C{H}_{3}COONa+H_{2}O....\Delta H=-13.2kcal$

$H^{\oplus }+\stackrel{\ominus }{O}H⟶H_{2}O;....\Delta H=-13.7Kcal$

$-13.7=-13.2-\left(\Delta H_{C{H}_{3}COOH}\right)$

$\Delta H_{C{H}_{3}COOH}=13.7-(-13.2)=-0.5$ kCal

Thus, the heat of dissociation for Acetic acid= -0.5 kcal

$HCl+N{H}_{4}OH⟶N{H}_{4}Cl+H_{2}O;\Delta H=-12.27kcal$

$-12.2=-13.2-\left(\Delta H_{N{H}_{4}OH}\right)$

$\Delta H_{N{H}_{4}OH}=-13.7-(-12.27)=-1.43$ kCal

Thus, the heat of dissociation for NH${}_4$OH$=-$1.43 cal

Above are enthalpies of formation. Negative of the above values give heat of dissociation.

Hence, the correct option is $B$