SATHEE: Thermodynamics and Thermochemistry 2 Question 1

Thermodynamics and Thermochemistry 2 Question 1

1. Enthalpy of sublimation of iodine is $24 cal g^{- 1}$ at $200^{\circ} C$ . If specific heat of $\mathrm{I}{2}(s) $a n d$ \mathrm{I}{2} $(v a p .) a r e 0.055 a n d$ 0.031 \mathrm{cal} \mathrm{g}^{-1} \mathrm{~K}^{-1} $r e s p e c t i v e l y, t h e n e n t h a l p y o f s u b l i m a t i o n o f i o d i n e a t$ 250^{\circ} \mathrm{C} $i n$ \mathrm{cal} \mathrm{g}^{-1}$ is

(2019 Main, 12 April I)

(a) 2.85

(b) 5.7

(d) 11.4

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Solution:

Key Idea When $q$ is the amount of heat involved in a system then at constant pressure

$q = q_{p} and C_{p} Δ T = Δ H$

Given reaction :

$$ \mathrm{I}{2}(s) \longrightarrow \mathrm{I}{2}(g) $$

Specific heat of $I_{2} (s) = 0.055 cal g^{- 1} K^{- 1}$ .

Specific heat of $I_{2} (vap) = 0.031 cal g^{- 1} K^{- 1}$ .

Enthalpy $(H_{1})$ of sublimation of iodine $= 24 cal g^{- 1}$

If $q$ is the amount of heat involved in a system then at constant pressure $q = q_{p}$ and

$\begin{aligned} Δ H & = C_{p} Δ T H_{2} - H_{1} & = C_{p} (T_{2} - T_{1}) H_{2} & = H_{1} + Δ C_{p} (T_{2} - T_{1}) H_{2} & = 24 + (0.031 - 0.055) (250 - 200) H_{2} & = 24 + (- 0.024) (50) = 24 - 1.2 = 22.8 cal / g \end{aligned}$

Thus, the enthalpy of sublimation of iodine at $250^{\circ} C$ is $22.8 cal / g$ .

Thermodynamics and Thermochemistry 2 Question 1

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Thermodynamics and Thermochemistry 2 Question 1

1. Enthalpy of sublimation of iodine is 24calg−1 at 200∘C. If specific heat of $\mathrm{I}{2}(s)and\mathrm{I}{2}(vap.)are0.055and0.031 \mathrm{cal} \mathrm{g}^{-1} \mathrm{~K}^{-1}respectively,thenenthalpyofsublimationofiodineat250^{\circ} \mathrm{C}in\mathrm{cal} \mathrm{g}^{-1}$ is

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

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