2
$\begingroup$

One of the back exercises (13(C)) from the book for practice provided at my school asked whether the following relation is true:

$\Delta_\mathrm{f}H^\circ(\ce{I2,g}) = \Delta_\mathrm{sub}H^\circ[\ce{I2,s}]$ at $\pu{25 °C}.$

I know how to solve the question, but the answer would depend on the meaning of $\Delta_\mathrm{sub}H^\circ[\ce{I2,s}].$ Which of the following reaction does the symbol $\Delta_\mathrm{sub}H^\circ[\ce{I2,s}]$ represent?

$$ \begin{align} \ce{I2(g) &-> I2(s)}\tag{a}\\ \ce{I2(s) &-> I2(g)}\tag{b} \end{align} $$

Improve this question
$\endgroup$
4
  • $\begingroup$ So Chaitanya Garg's answer is correct. And the answer is (b). $\endgroup$
    – Maurice
    Feb 28 '21 at 9:59
  • 1
    $\begingroup$ Sublimation by definition refers to phase transition from solid to gas (no textbook or reference required I would think, but see: en.wikipedia.org/wiki/Enthalpy_of_sublimation). $\endgroup$
    – Buck Thorn ♦
    Feb 28 '21 at 10:28
  • $\begingroup$ @BuckThorn I'm more interested in why they used square brackets. Have you ever seen them used like this for the thermodynamic quantities? Or am I looking for the meaning where there is none? $\endgroup$
    – andselisk ♦
    Feb 28 '21 at 10:31
  • $\begingroup$ @andselisk Rather odd to have different types of brackets on both sides of the OPs equation. Typo? $\endgroup$
    – Buck Thorn ♦
    Feb 28 '21 at 10:36
2
$\begingroup$

$\ ΔH_{sub}(I_2,s)$ can be interpreted as the change in enthalpy when one mole of solid $\ I_2$ converts to gaseous $\ I_2$ at a constant temperature. So the answer should be reaction (b).

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Could you please provide with a reference text that says so. $\endgroup$
    – Aseem Mittal
    Feb 28 '21 at 9:57
  • $\begingroup$ @Aseem Mittal, Pg No. 50 Atkins' Physical Chemistry by Peter Atkins and also Pg. No. 165 of NCERT textbook for class 11th. $\endgroup$
    – Chaitanya Garg
    Feb 28 '21 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.