Question

The specific heats of iodine vapour and solid are $0.031cal{K}^{-1}{g}^{-1}$ and $0.055cal{K}^{-1}{g}^{-1}$ respectively. If heat of sublimation of iodine is 24 cal/g at ${200}^{\circ }C$, calculate its value at ${250}^{\circ }C$.

• A. $5.7cal{g}^{-1}$
• B. $11.4cal{g}^{-1}$
• C. $22.8cal{g}^{-1}$
• D. $45.6cal{g}^{-1}$

Answer 1

The correct option is C $22.8cal{g}^{-1}$

Solid $I_2$ is converted into $I_2$ vapour
$I_2\left(s\right)\to I_2\left(g\right)\Delta H=24cal/gat{200}^{\circ }C\Delta C_p=\left(C_p\right)I_2\left(g\right)-\left(C_p\right)I_2\left(s\right)=0.031-0.055=-0.024cal{g}^{-1}{K}^{-1}$
From Kirchhoff's equation,
$\Delta H_{{250}^{\circ }}-\Delta H_{{200}^{\circ }C}=\Delta C_p(T_{{250}^{\circ }}-T_{{200}^{\circ }})\Delta H_{{250}^{\circ }C}=24-0.024(523-473)=24-1.2=22.8cal{g}^{-1}$