# Meaning of the symbol “$\Delta_\mathrm{sub}H^\circ[\ce{I2,s}]$”

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}

• So Chaitanya Garg's answer is correct. And the answer is (b). Feb 28 '21 at 9:59
• 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). Feb 28 '21 at 10:28
• @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? Feb 28 '21 at 10:31
• @andselisk Rather odd to have different types of brackets on both sides of the OPs equation. Typo? Feb 28 '21 at 10:36

$$\ ΔH_{sub}(I_2,s)$$ 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).