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JEE Main 201912 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Main 2019 · 12 April · Morning Shift · Chemistry · Thermodynamics (C)

Enthalpy of sublimation of iodine is 24cal g ^ - 1 at 200 ° C . If...

Enthalpy of sublimation of iodine is

24cal{g}^{- 1}

at

200 ° C

. If specific heat of

{I}_{2}

(s)

and {I}_{2}

(vap) are

0.055

and

0.031cal{g}^{- 1}{K}^{- 1}

respectively, then enthalpy of sublimation of iodine at

250 ° C

in

cal{g}^{- 1}

is:

Options

A $11.4$
B $2.85$
C $22.8$
D $5.7$

Correct answer

C. $22.8$

Step-by-step solution

{I}_{2}\left(s\right) \to {I}_{2}\left(g\right)

{\left(\DeltaH\right)}_{{250}^{o}C} - {\left(\DeltaH\right)}_{{200}^{o}C} = \Delta{C}_{pm}\left({T}_{2} - {T}_{1}\right)

{\left(\DeltaH\right)}_{{250}^{o}C} - 24 = \left(0.013 - 0.055\right) \times 50

{\left(\DeltaH\right)}_{{250}^{o}C} - 24 = - 1.2

{( \DeltaH )}_{{250}^{o}C} = 22.8J / gm

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