JEE PYQ: Thermodynamics Question 28
Question 28 - 2019 (12 Apr Shift 1)
Enthalpy of sublimation of iodine is 24 cal g$^{-1}$ at $200^\circ$C. If specific heat of $\text{I}_2(s)$ and $\text{I}_2(\text{vap})$ are 0.055 and 0.031 cal g$^{-1}$K$^{-1}$ respectively, then enthalpy of sublimation of iodine at $250^\circ$C in cal g$^{-1}$ is:
(1) 2.85
(2) 5.7
(3) 22.8
(4) 11.4
Show Answer
Answer: (3)
Solution
$\text{I}2(s) \to \text{I}2(g)$. Heat of reaction depends upon temperature, as given by Kirchoff’s equation: $\Delta H{T_2} = \Delta H{T_1} + \int_{T_1}^{T_2}\Delta C_p dT$. $\Delta C_p = C_p \text{ of product} - C_p \text{ of reactant} = 0.031 - 0.055 = -0.024$ cal/g. $\Delta H_{T_2} - \Delta H_{T_1} = \Delta C_p(T_2 - T_1) = -0.024(523 - 473)$. $\Delta H_{250^\circ\text{C}} = 24 - 50 \times 0.024 = 22.8$ cal/g.
BETA
