Which of the following reactions is accompanied by a decrease in entropy?

A decrease in entropy ($\Delta S < 0$) typically occurs when the number of moles of gas decreases.

• A. $\text{N}_2\text{O}_{4(g)} \rightleftharpoons 2\text{NO}_{2(g)}$ Reactant moles of gas = 1 Product moles of gas = 2 Change in moles of gas ($\Delta n_{gas}$) = $2 - 1 = +1$. Entropy increases.

• B. $\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)}$ Reactant moles of gas = $1 + 3 = 4$ Product moles of gas = 2 Change in moles of gas ($\Delta n_{gas}$) = $2 - 4 = -2$. Entropy decreases.

• C. $\text{CaCO}_{3(s)} \rightleftharpoons \text{CaO}_{(s)} + \text{CO}_{2(g)}$ Reactant moles of gas = 0 Product moles of gas = 1 Change in moles of gas ($\Delta n_{gas}$) = $1 - 0 = +1$. Entropy increases.

• D. $2\text{N}_2\text{H}_{4(l)} + \text{N}_2\text{O}_{4(l)} \rightleftharpoons 3\text{N}_{2(g)} + 4\text{H}_2\text{O}_{(l)}$ Reactant moles of gas = 0 (both are liquids) Product moles of gas = 3 Change in moles of gas ($\Delta n_{gas}$) = $3 - 0 = +3$. Entropy increases.

The reaction accompanied by a decrease in entropy is B.

The final answer is $\boxed{\text{B}}$

Given atomic masses: $\text{Cu} = 64$, $\text{S} = 32$, $\text{O} = 16$.

$$\text{Molar mass of } \text{CuSO}_4 = \text{Cu} + \text{S} + (4 \times \text{O})$$ $$\text{Molar mass of } \text{CuSO}_4 = 64 + 32 + (4 \times 16)$$ $$\text{Molar mass of } \text{CuSO}_4 = 64 + 32 + 64 = 160 \text{ g/mol}$$

Given mass of $\text{CuSO}_4 = 32 \text{ g}$.

$$\text{Moles of } \text{CuSO}_4 = \frac{\text{Mass}}{\text{Molar mass}}$$ $$\text{Moles of } \text{CuSO}_4 = \frac{32 \text{ g}}{160 \text{ g/mol}} = 0.2 \text{ mol}$$

Heat generated = $13.0 \text{ kJ}$.

$$\text{Heat of solution} = \frac{\text{Heat generated}}{\text{Moles of solute}}$$ $$\text{Heat of solution} = \frac{13.0 \text{ kJ}}{0.2 \text{ mol}} = 65.0 \text{ kJ/mol}$$

Since heat was generated, the process is exothermic, and the enthalpy change would be $-65.0 \text{ kJ/mol}$. However, the options provide the magnitude.

The final answer is $\boxed{\text{B. 65.0 KJmol}^{-1}}$