Question

The heat of formation of one mole of HI from hydrogen and iodine vapour at ${25}^{\circ }C$ is $-$8000 Cal.
$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{I}_2\left(g\right)\to HI\left(g\right);\Delta H=-8000Cal$
The heat of formation of one mole of HI at ${10}^{\circ }C.$ is:
Given : $C_P=6.5+0.0017$ T for hydrogen (g) ( Cal / mol / K )
$C_P=6.5+0.0031$ T for iodine (g) ( Cal / mol / K )
$C_P=6.5+0.0016$ T for HI (g) ( Cal / mol / K )

• A. $-$8934.5 Cal
• B. $-$7996.5 Cal
• C. $-$5886.5 Cal
• D. $-$3866.5 Cal

Answer 1

The correct option is B $-$7996.5 Cal

$\frac{1}{2}{H}_2+\frac{1}{2}{I}_2\to HI$
$\Delta C_p=C_p\left(HI\right)-[\frac{1}{2}{C}_p\left(H_2\right)+\frac{1}{2}{C}_p\left(I_2\right)]$
$=(6.5+0.0016T)-[\frac{1}{2}(6.5+0.0017T)+\frac{1}{2}(6.5+0.031T)]$
$=0.0016T-\frac{0.0017T}{2}-\frac{0.0031T}{2}=-0.0008TCal/K$
$T_1=273+10=283K$ $T_2=273+25=298K$
${\Delta }_f{H}_{298}=-8000Cal$
${\Delta }_f{H}_{298}={\Delta }_f{H}_{283}+{\int }_{283}^{298}\Delta C_{p}dT$
$\Rightarrow -8000={\Delta }_f{H}_{283}+{\int }_{283}^{298}(-0.0008)dT$
$\Rightarrow {\Delta }_f{H}_{283}=-8000+{\left[\frac{0.0008T^2}{2}\right]}_{283}^{298}=-8000+0.0004[{298}^2-{283}^2]=-8000+3.486$
$\therefore {\Delta }_f{H}_{283}=-7996.5Cal$