$H A + O H^{-} ⟶ H_{2} O + A^{-} + q_{1} k J$
$H^{+} + O H^{-} ⟶ H_{2} O + q_{2} k J$
The enthalpy of dissociation of $H A$ is:

(q_{1} + q_{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(q_{1} - q_{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(q_{2} - q_{1})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- (q_{1} + q_{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $(q_{2} - q_{1})$
Solution:- (C) $(q_{2} - q_{1}) k J$

$H A + {O H}^{-} ⟶ H_{2} O + A^{-} + q_{1} k J . . . . . (1)$

$H^{+} + {O H}^{-} ⟶ H_{2} O + q_{2} k J$

$H_{2} O ⟶ H^{+} + {O H}^{-} - q_{2} k J . . . . . (2)$

Adding ${e q}^{n} (1) & (2)$ , we have

$H A + {O H}^{-} + H_{2} O ⟶ H_{2} O + A^{-} + H^{+} + {O H}^{-} + (q_{1} - q_{2}) k J$

$H A ⟶ H^{+} + {O H}^{-}; Δ H = - (q_{1} - q_{2}) k J$

Hence the enthalpy of dissociation of $H A$ is $(q_{2} - q_{1}) k J$ .

Suggest Corrections

Similar questions

C H_{3} C O O H + O H^{-} ⇋ C H_{3} C O O^{-} + H_{2} O + q_{1}

H^{+} + O H^{-} ⇋ H_{2} O + q_{2}

C H_{3} C O O H ⇋ C H_{3} C O O^{-} + H^{+}

C H_{3} C O O H + O H^{-} \to C H_{3} C O O^{-} + H_{2} O + q_{1}

H^{+} + O H^{-} \to H_{2} O + q_{2}

C H_{3} C O O H \to C H_{3} C O O^{-} + H^{+}

q_{1} + q_{2} = q

q_{1}

q_{2}

\frac{q_{1}}{q_{2}}

Consider the situation shown in figure. What are the signs of $q_{1} a n d q_{2}$ ?

Q_{1}, Q_{2}

Q_{3}

Join BYJU'S Learning Program