Enthalpy of sublimation of iodine is $40 c a l g^{- 1} a t 200^{0} C$ . If specific heat of $I_{2} (s) a n d I_{2} (v a p)$ are 0.055 and 0.031 cal $g^{- 1} K^{- 1}$ respectively, then enthalpy of sublimation of iodine at $250^{0} C$ in cal $g^{- 1}$ is ______ [nearest integer]

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 39.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$Δ C_{p} = \frac{H_{2} - H_{2}}{T_{2} - T_{1}} I_{2 (s)} \to I_{2 (g)}, Δ I_{1} = 40 c a l / g a 200^{0} C Δ H_{2} = Δ H_{1} + C_{P_{r}} (T_{2} - T_{1}) = 40 + (0.031 - 0.055) \times (250 - 200) = 40 - 1.2 = 38.8 c a l / g r a m$

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!