Question

The molar heat capacities of Iodine vapour and solid are $7.8$ and $14cal/mol$ respective enthalpy of sublimation of iodine is $6096cal/mol$ at ${200}^{\circ }C$, then what is $\Delta H$ the value at ${250}^{\circ }C$ in cal/mol.

• A. $5360$
• B. $4740$
• C. $6406$
• D. none of these

Answer 1

Answer: (A)

$5360$

$\Delta C_p=\frac{\Delta H_2-\Delta H_1}{T_2-T_1}$

$14-7.8=\Delta H_2-609650$

$6.2\times 50=\Delta H_2-6096$

$6.2\times 50+6096=\Delta H_2$

$\Delta H_2=310+6096=6406$ cal/mol