Question

The dissociation constant of ${NH}_{4}OH$ at ${25}^{o}C$ is $K_b=X\times {10}^{-6}$ , if $\Delta H^o$ and $\Delta S^o$ for the given changes are as follows: ${NH}_4+H^{+}\rightleftharpoons {NH}_4^{+};\Delta H^o=-52.2kJ{mol}^{-1};\Delta S^o=+1.67J{K}^{-1}{mol}^{-1}$ $H_{2}O\rightleftharpoons H^{+}+O{H}^{-};\Delta H^o=56.6kJ{mol}^{-1};\Delta S^o=-78.2J{K}^{-1}{mol}^{-1}$. Value of X is=5Y+2. What is the value of Y?

Answer 1

${NH}_{4}OH+H^{+}\rightleftharpoons {NH}_4^{+};\Delta H^o=-52.2kJ{mol}^{-1}$
Adding $H_{2}O\rightleftharpoons H^{+}+O{H}^{-};\Delta H^o=+56.6kJ{mol}^{-1}$
$\therefore$ ${NH}_3+H_{2}O\rightleftharpoons {NH}_4^{+}+O{H}^{-};\Delta H^o=+4.4kJ{mol}^{-1}$
Similarly, $\Delta S^o$ for the change $=-76.53J{K}^{-1}$ ${mol}^{-1}$
or for the change:
${NH}_3+H_{2}O\rightleftharpoons {NH}_4^{+}+O{H}^{-};\Delta H^o=4.4kJ{mol}^{-1}$
and $\Delta S^o=-76.53J{K}^{-1}$ ${mol}^{-1}$
Now, we have $\Delta G^o=\Delta H^o-T\Delta S^o$
$\therefore \Delta G^o=4.4-(-76.53\times {10}^{-3})\times 298$ or
$=27.21kJ$ ${mol}^{-1}$
Also, $\Delta G^o=-2.303RT$ $\log K_b$
$27.21=-2.303\times 8.314\times {10}^{-3}\times 298\times \log K_b$
$\therefore$ $K_b=1.7\times {10}^{-5}$

so Y=3.