Question

Which of the following graphs correctly represents the variation of thermodynamic properties of Haber's process?

• A.
• B.
• C.
• D.

Hint:

Haber process is exothermic with negative entropy change; use $-\Delta G/T = -\Delta H/T + \Delta S$.

Answer 1

The Correct Option is A

Step 1: Signs for the Haber reaction: $\Delta H^\theta$ is negative (about -92 kJ per mole of reaction) and $\Delta S^\theta$ is negative (4 mol of gas become 2 mol).
Step 2: So the curve for $\Delta H^\theta$ and $\Delta S^\theta$ lie below the axis and are nearly flat, while $-\Delta H^\theta/T$ is positive and decreases like $1/T$.
Step 3: $-\Delta G^\theta/T = -\Delta H^\theta/T + \Delta S^\theta$ is the $-\Delta H^\theta/T$ curve shifted down by the (negative) $\Delta S^\theta$: it starts positive, lies below $-\Delta H^\theta/T$ and cuts the axis where $\Delta G^\theta = 0$.
Step 4: This matches graph (1) only: option (1).