Question

Consider the table below, showing the formation of the hydrogen halides, with their accompanying $\Delta H^{\circ }$ and $\Delta S^{\circ }$ values. Assume each reaction occurs at constant pressure.

	Reaction	$\Delta H^{\circ }$	$\Delta S^{\circ }$
A	$H_2\left(g\right)+F_2\left(g\right)\to 2HF\left(g\right)$	$-273kJ/mol$	$+174J/molK$
B	$H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$	$-92kJ/mol$	$+187J/molK$
C	$H_2\left(g\right)+B{r}_2\left(l\right)\to 2HBr\left(g\right)$	$-36kJ/mol$	$+199J/molK$
D	$H_2\left(g\right)+I_2\left(s\right)\to 2HI\left(g\right)$	$+27kJ/mol$	$+207J/molK$

Which of the four statements is true, based on the information in the table?

• A. Of the four, only Reaction A is spontaneous at all temperatures.
• B. Reaction A will have the greatest rate of reaction.
• C. Hydrogen fluoride, $HF$, has the greatest bond dissociation energy of the for hydrogen halides.
• D. Of the four, only Reaction D is spontaneous at all temperatures.

Answer 1

Answer: (C)

Hydrogen fluoride, $HF$, has the greatest bond dissociation energy of the for hydrogen halides.

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

For a spontaneous reaction, $\Delta G$ should be negative. For reactions (A), (B) and (C), $\Delta G$ is always negative. Therefore, reactions (A), (B) and (C) are spontaneous.

$\Delta H=$ Energy of reactants - Energy of products

Therefore, higher the negative value of $\Delta H$, higher will be the bond dissociation energy of the product making the product more stable.

${2HF}_{\left(g\right)}\to H_2\left(g\right)+F_2\left(g\right);\Delta H=-273kJ/mol$

$2HCl\left(g\right)\to H_2\left(g\right)+{Cl}_2\left(g\right);\Delta H=-92kJ/mol$

$2HBr\left(g\right)\to H_2\left(g\right)+{Br}_2\left(g\right);\Delta H=-36kJ/mol$

$2HI\left(g\right)\to H_2\left(g\right)+I_2\left(g\right);\Delta H=+27kJ/mol$

Therefore, $HF$ has the greatest bond dissociation energy of the four hydrogen halides.