Question

The molar entropies of $HI\left(g\right),H\left(g\right)$ and $I\left(g\right)$ at $298K$ are $206.5,114.6$, and $180.7Jmol{e}^{-1}{K}^{-1}$ respectively. Using the $\Delta G^0$ given below,Calculate the bond energy of HI.

$H\left(g\right)\to H\left(g\right)+I\left(g\right)$; $\Delta G^0=271.8kJ$

• A. 282.4 kJ/mole
• B. 298.3 kJ/mole
• C. 32.17kJ/mole
• D. 100 kJ/mole

Answer 1

The correct option is B

298.3 kJ/mole

$H{I}_{\left(g\right)}\to H{I}_{\left(g\right)}+I_{\left(g\right)}$

$\Delta G^0=\Delta G-T\Delta S^0$

$\Delta S^0=\sum S^{0}products-\sum S^{0}Reactants$

= $(114.6+180.7)-\left(206.5\right)=295.3-206.5=88.8$

$\Delta G^0=\Delta H-T\Delta S$

$\Delta H=\Delta G^0+T\Delta S=27.18+298\times 88.8=271.8+26462.8Joules$

= $271.8+26.4628=298.3kJ/mole$