Question

It is said that ∆S is positive as randomness is increasing from diamond to graphite as in diamond molecules are tightly packed together but we know that ∆S=∆H/T for phase transition and that conversion of diamond to graphite is exothermic which indicates ∆H as negative so shouldn't ∆S be negative? (Correct me if I am wrong.)

Answer 1

The equation ΔS = ΔH/T is specific to isothermal phase changes and does not apply to the allotrope change from diamond to graphite. The increase in entropy (ΔS > 0) is due to the higher disorder and number of microstates in graphite compared to the tightly packed structure of diamond.

Step-by-step explanation:

You seem to have a misunderstanding regarding the equation ΔS = ΔH/T. This is not the general equation for entropy change, but rather a specific case for phase transitions. In the case of the conversion from diamond to graphite, we're dealing with a change in allotrope, not a phase change. The main factor we are looking at here is the change in the degree of order as we move from a highly ordered structure (diamond) to a less ordered structure (graphite), which is why we see an increase in entropy, or ΔS > 0.

The formula you're referring to, ΔS = ΔH/T, applies to reversible isothermal phase transitions, such as melting or boiling. Even though the conversion of diamond to graphite is exothermic (ΔH < 0), the sign of ΔS does not directly depend on ΔH in this context. Instead, entropy is related to the amount of disorder in the system. Graphite, with its less rigid structure and higher number of microstates, has greater entropy than diamond. Therefore, the conversion of diamond to graphite results in an increase in entropy despite the negative ΔH.