Question

question 5 which of the following reactions will have positive change in entropy? caco₃ (s) → cao (s) + co₂ (g) h₂o (g) → h₂o (l) n₂ (g) + 3 h₂ (g) → nh₃ (g) ch₄ (g) + 2 o₂ (g) → co₂ (g) + h₂o (l)

Explanation:

To determine which reaction has a positive change in entropy ($\Delta S > 0$), we analyze the change in the number of moles of gas and the phase changes (since entropy generally follows the order: gas > liquid > solid).

Step 1: Analyze $CaCO_3 (s)

ightarrow CaO (s) + CO_2 (g)$

• Reactants: 1 mole of solid.
• Products: 1 mole of solid + 1 mole of gas.
• The production of a gas from a solid increases the disorder (entropy), so $\Delta S$ is positive.

Step 2: Analyze $H_2O (g)

ightarrow H_2O (l)$

• Reactant: 1 mole of gas.
• Product: 1 mole of liquid.
• A gas to liquid phase change decreases disorder (entropy), so $\Delta S$ is negative.

Step 3: Analyze $N_2 (g) + 3 H_2 (g)

ightarrow NH_3 (g)$

• Reactants: 1 + 3 = 4 moles of gas.
• Products: 2 moles of gas (since the coefficient of $NH_3$ is 2? Wait, no, the balanced equation is $N_2 + 3H_2

ightarrow 2NH_3$. So reactants: 4 moles of gas; products: 2 moles of gas.

• A decrease in the number of moles of gas decreases disorder (entropy), so $\Delta S$ is negative.

Step 4: Analyze $CH_4 (g) + 2 O_2 (g)

ightarrow CO_2 (g) + H_2O (l)$

• Reactants: 1 + 2 = 3 moles of gas.
• Products: 1 mole of gas + 1 mole of liquid.
• A decrease in the number of moles of gas (and a gas to liquid change) decreases disorder (entropy), so $\Delta S$ is negative.

Answer:

The reaction with a positive change in entropy is $\boldsymbol{CaCO_3 (s)
ightarrow CaO (s) + CO_2 (g)}$ (the first option).