Question

At 300 K and constant pressure, the following data is obtained for the reaction: \[ 4M(s) + 3O_2(g) \rightarrow 2M_2O_3(s) \] Given: \[ \Delta H^\circ = -1548 \, \text{kJ mol}^{-1}, \quad \Delta S_{\text{sys}} = -550 \, \text{J K}^{-1}\text{mol}^{-1} \] What is the value of $\Delta S_{\text{surr}}$ in J K$^{-1}$ mol$^{-1}$?

• A. $+$4644
• B. $+$5160
• C. $-$4644
• D. $-$5160

Hint:

For any process: \(\Delta S_{\text{surr}} = -\Delta H/T\) at constant pressure and temperature.

Answer 1

The Correct Option is B

Step 1: Think about where the heat from an exothermic reaction goes.
The reaction releases heat because $\Delta H^{\circ}$ is negative. At constant pressure this heat flows out of the system and into the surroundings, so the surroundings gain heat and their entropy goes up.

Step 2: Write the surroundings' entropy change in terms of the heat they receive.
\[ \Delta S_{surr} = \frac{q_{surr}}{T} = \frac{-\Delta H_{sys}}{T} \]
The minus sign carries over the fact that heat lost by the system is heat gained by the surroundings.

Step 3: Keep the enthalpy in kJ for now and substitute directly.
\[ \Delta S_{surr} = \frac{-(-1548)}{300} = \frac{1548}{300} \, \text{kJ K}^{-1}\text{mol}^{-1} \]

Step 4: Do the division first.
\[ \Delta S_{surr} = 5.16 \, \text{kJ K}^{-1}\text{mol}^{-1} \]

Step 5: Convert to joules only at the last step.
\[ 5.16 \, \text{kJ K}^{-1}\text{mol}^{-1} \times 1000 = 5160 \, \text{J K}^{-1}\text{mol}^{-1} \]

Step 6: Check the sign makes physical sense.
A positive $\Delta S_{surr}$ fits an exothermic reaction, since the surroundings should gain disorder when they absorb heat. This rules out the negative options (3) and (4) immediately, and the correct magnitude rules out (1).

Final Answer:
\[ \boxed{+5160 \, \text{J K}^{-1}\text{mol}^{-1}} \]