Question

The specific heats of iodine vapour and solid are $0.031$ and $0.055cal{K}^{-1}{g}^{-1}$ respectively. If heat of sublimation of iodine is $24cal{g}^{-1}$ at ${200}^{o}C$, find 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 in $I_2$ vapour,
The reaction will be,
$I_2\left(s\right)\to I_2\left(g\right)△H=24cal{g}^{-1}at{200}^{o}C$

$△C_p=(C_p{)}_{product}-(C_p{)}_{reactant}$
$△C_p=\left(C_p\right)I_2\left(g\right)-\left(C_p\right)I_2\left(s\right)$
$=0.031-0.055=-0.024cal{K}^{-1}{g}^{-1}$

From Kirchhoff's equation
$△H_{{250}^{o}C}-△H_{{200}^{o}C}=△C_p(T_{{250}^{o}C}-T_{{200}^{o}C})$
putting the values,
$△H_{{250}^{o}C}=24-0.024(523-473)$
$=24-1.2$
$\Rightarrow △H_{{250}^{o}C}=22.8cal{g}^{-1}$