Question

From the following data, mark the option(s) where $\Delta$H is correctly written for the given reaction.

Given : $H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to H_{2}O\left(\ell \right);\Delta H=-57.3kJ$

$\Delta H_{\text{solution}}\text{of}\text{HA(g)}=-70.7\text{kJ/mol}$

$\Delta H_{\text{solution}}\text{of}\text{BOH(g)}=+20\text{kJ/mol}$

$\Delta H_{\text{ionization}}\text{of}HA=15\text{kJ/mol}$ and BOH is a strong base.

$\begin{array}{ccc}& \text{Reaction}& \Delta H_r\text{(kJ/mol)}\\ \text{(a)}& \text{HA(aq)}+\text{BOH(aq)}\to \text{BA(aq)}+H_{2}O& -42.3\\ \text{(b)}& \text{HA(g)}+\text{BOH(g)}\to \text{BA(aq)}+H_{2}O& -93\\ \text{(c)}& \text{HA(g)}\to H^{+}\left(aq\right)+A^{-}\left(aq\right)& -55.7\\ \text{(d)}& B^{+}\text{(aq)}+O{H}^{-}\text{(aq)}\to \text{BOH(aq)}& -20\end{array}$

• A. $\begin{array}{ccc}& \text{Reaction}& \Delta H_r\text{(kj/mol)}\\ \text{(a)}& \text{HA(aq)}+\text{BOH(aq)}\to \text{BA(aq)}+H_{2}O& -42.3\end{array}$
• B. $\begin{array}{ccc}& \text{Reaction}& \Delta H_r\text{(kj/mol)}\\ \text{(b)}& \text{HA(g)}+\text{BOH(g)}\to \text{BA(aq)}+H_{2}O& -93\end{array}$
• C. $\begin{array}{ccc}& \text{Reaction}& \Delta H_r\text{(kj/mol)}\\ \text{(c)}& \text{HA(g)}\to H^{+}\left(aq\right)+A^{-}\left(aq\right)& -55.7\end{array}$
• D. $\begin{array}{ccc}& \text{Reaction}& \Delta H_r\text{(kj/mol)}\\ \text{(d)}& B^{+}\text{(aq)}+O{H}^{-}\text{(aq)}\to \text{BOH(aq)}& -20\end{array}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A

$\begin{array}{ccc}& \text{Reaction}& \Delta H_r\text{(kj/mol)}\\ \text{(a)}& \text{HA(aq)}+\text{BOH(aq)}\to \text{BA(aq)}+H_{2}O& -42.3\end{array}$

B

$\begin{array}{ccc}& \text{Reaction}& \Delta H_r\text{(kj/mol)}\\ \text{(b)}& \text{HA(g)}+\text{BOH(g)}\to \text{BA(aq)}+H_{2}O& -93\end{array}$

C

$\begin{array}{ccc}& \text{Reaction}& \Delta H_r\text{(kj/mol)}\\ \text{(c)}& \text{HA(g)}\to H^{+}\left(aq\right)+A^{-}\left(aq\right)& -55.7\end{array}$

$H{A}_{\left(g\right)}+aq\to HA\left(aq\right)\Delta H=-70.7....\left(1\right)$

$BO{H}_{\left(g\right)}+aq\to BOH\left(aq\right)\Delta H=20....\left(2\right)$

$H{A}_{\left(g\right)}+BO{H}_{\left(g\right)}\to HA\left(aq\right)+BOH\left(aq\right)\Delta H=-50.7....\left(3\right)$

$HA\left(aq\right)+BOH\left(aq\right)\to BA\left(aq\right)+H_{2}O\Delta H=(-57.3+15)=-42.3....\left(4\right)$

Remember, the formation of water, $H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to H_{2}O\left(\ell \right);\Delta H=-57.3kJ$ is hidden in the above equation. When you mix the aq acid and the aq base, you get the salt BA and water forms. The ionisation of HA requires a heat input of 15 kJ/mol and the formation of water releases 57.3 kJ/mol. That's what the above reaction is telling us.

The rest of the steps are self explanatory, just use Hess Law and match the relevant options.

$\left(3\right)+\left(4\right)$

$H{A}_{\left(g\right)}+BO{H}_{\left(g\right)}\to BA\left(aq\right)+H_{2}O\Delta H=-50.7-42.3=-93$

$HA\left(aq\right)\to H^{+}\left(aq\right)+A^{-}\left(aq\right)\to \Delta H=15....\left(5\right)$

$\left(1\right)+\left(5\right)$

$HA\left(g\right)\to H^{+}\left(aq\right)+A^{-}\left(aq\right)\Delta H=-70.7+15=-55.7$