The mathematical expression for the standard enthalpy of sublimation is given by

Chemistry · JEE Advanced · NTA Exams — Chemical Thermodynamics

1. ∆Hº (sublimation) = ∆Hº (fusion) – 2∆Hº (vaporization)
2. ∆Hº (sublimation) = ∆Hº (fusion) – ∆Hº (vaporization)
3. ∆Hº (sublimation) = ∆Hº (fusion) + ∆Hº (vaporization)
4. ∆Hº (sublimation) = ∆Hº (combustion) + ∆Hº (dissociation)

Answer

(C) ∆ Hº (sublimation) = ∆ Hº (fusion) + ∆ Hº (vaporization)

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Related practice questions

• Calculate the enthalpy change when 1.0 g of water is frozen at 0 0 C ( H _ f ^ 0 = 1.435 kcal mol –1 )
• 1 mole of an ideal gas A (C v,m = 3 R) and 2 mole of an ideal gas B ( C _ v , ~m = 3 2 R ) are taken in a…
• When heat is supplied to an ideal gas in an isothermal process, the
• The enthalpy of reaction CH 4 (g) + O 2 (g) → CH 3 OH( L ) is negative. If enthalpy of combustion of CH 4 and…
• 10 mole of an ideal gas is heated at constant pressure of one atmosphere from 27 o C to 127 o C.If C_ v, m…
• ∆ r H 0 for solid-to-liquid transition for proteins A and B are 2.73 kcal mol –1 and 3.03 kcal mol –1 . The…
• A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of…
• S = H ~T holds good for

More Chemical Thermodynamics questions · Browse all practice questions