Question

Solve for the standard entropy change (ΔS⁰) with each reaction below. As practice, try to predict what the sign would be before you solve it and see if it matches up.

a. N2(g) + 3 H2(g) ⇌ 2NH3(g)
b. NH4Cl(s) ⇌ NH3(g) + HCl(g)
c. CO(g) + 2H2(g) ⇌ CH3OH(l)
d. Li3N(s) + 3H2O(l) ⇌ 3 LiOH(aq) + NH3(g)

Answer 1

Step-by-step explanation:

The term
`\Delta S` means change in entropy. As entropy means the measure of randomness. This means that more randomly molecules of a substance are moving more will be its entropy.

If value of
`\Delta S` is negative then it means there is decease in entropy. When value of
`\Delta S` is positive then it means there is increase in entropy.

In solids, molecules are closer to each other. So, entropy is minimum. Liquids has more entropy than solids and gases has maximum entropy.

• In the reaction,
  `N_2(g) + 3H_2(g) \rightleftharpoons 2NH_(3)(g)`, total 4 moles of gases form 2 mole of a gas. This means there is decrease in number of moles. As a result, there will be decrease in entropy. So, sign of
  `\Delta S` is negative.

• In the reaction,
  `NH_(4)Cl(s) \rightleftharpoons NH_(3)(g) + HCl(g)`, total 1 mole of solid substance is forming total 2 moles of gases. That is, there is increase in entropy. So, sign of
  `\Delta S` is positive.

• In the reaction,
  `CO(g) + 2H_(2)(g) \rightleftharpoons CH_(3)OH(l)(g)`, total 2 moles of gases are forming total 1 mole of liquid methanol. That is, there is decrease in entropy. So, sign of
  `\Delta S` is negative.

• In the reaction,
  `Li_(3)N(s) + 3H_(2)O(l) \rightleftharpoons 3LiOH(aq) + NH_(3)(g)`, total 2 moles of substance is forming total 4 moles of substance. That is, there is increase in entropy so, sign of
  `\Delta S` is positive.