Several factors can dictate entropy in an equation.
These include:
1. Phase changes
⇒ When a solid turns to a liquid, the entropy increases as the particles have more freedom to move around and thus have a greater ability for 'disorder'. Same goes for a liquid turning to a gas. In a gas, the intermolecular forces are much weaker than that of a solid or liquid, allowing the particles more freedom.
So, going from a solid to liquid to gas increases entropy, and going the other way, from gas to liquid to solid, decreases entropy.
Example:
H₂O(l) -> H₂O(g)
This will have a positive entropy change, as the water molecules are becoming gaseous and thus have more freedom.
2. Dissolution
⇒ Similarly, breaking up particles of a solute when dissolving in a solvent will increase entropy as the particles are no longer bound together.
So, dissolving a solute will increase entropy.
Example:
NaCl(s) -> NaCl(aq)
This will have a positive entropy change, as the NaCl particles are more free after being separated.
3. Number of products and reactants
⇒ Generally, if you have more moles of products than reactants, if they are the same phase then entropy will increase. Note this is not necessarily true if you form a gas from two non-gas reactants, as the gas will still have more entropy.
4. Temperature
⇒ Increasing temperature will increase entropy as the particles have more kinetic energy and are then moving faster.
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l. 2N2(g) + O2(g) → 2N2O(g)
3 moles of gas are forming 2 moles of gas. The phase of products and reactants are the same, so since we have less moles of product than reactant, entropy will be negative.
II. CaCO3(s) → CaO(s) + CO2(g)
1 mole of solid is forming 1 mole of solid and 1 mole of gas. There is a phase change from solid to gas, and there are more moles of product than reactant, entropy will be positive.
III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
While 3 moles of reactant are forming only 2 moles product, we are forming a gas from non-gaseous reactants, so entropy will be positive regardless.