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DeltaS("says") "for" 4Fe((s)) + 3O(2) ra...

`DeltaS_("says") "for" 4Fe_((s)) + 3O_(2) rarr 2Fe_(2) O_(3(s)) " is" -550 J//k//mol` at 298K. If enthalpy change for same process is `-1600kJ//mol, DS_("total")` (in J/mol/K) is

A

`[(1600)/(298) xx 10^(3)] + 550 gt 0`

B

`550- [(1600)/(298)] lt0`

C

`[(1600)/(298) xx 10^(3)] -550 gt 0`

D

`[(1600+550)/(298)] gt 0`

Text Solution

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The correct Answer is:
To solve the problem, we need to calculate the total entropy change (ΔS_total) for the given reaction: **Given:** - Reaction: \( 4 \text{Fe}_{(s)} + 3 \text{O}_{2(g)} \rightarrow 2 \text{Fe}_2\text{O}_{3(s)} \) - ΔS_system = -550 J/K·mol - ΔH = -1600 kJ/mol - Temperature (T) = 298 K ### Step-by-Step Solution: **Step 1: Convert ΔH from kJ to J.** \[ \Delta H = -1600 \text{ kJ/mol} \times 1000 \text{ J/kJ} = -1600000 \text{ J/mol} \] **Step 2: Calculate ΔS_surroundings using the formula:** \[ \Delta S_{\text{surroundings}} = \frac{\Delta H}{T} \] Substituting the values: \[ \Delta S_{\text{surroundings}} = \frac{-1600000 \text{ J/mol}}{298 \text{ K}} \approx -5376.51 \text{ J/mol·K} \] **Step 3: Calculate ΔS_total using the formula:** \[ \Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} \] Substituting the values: \[ \Delta S_{\text{total}} = -550 \text{ J/mol·K} + (-5376.51 \text{ J/mol·K}) \] \[ \Delta S_{\text{total}} = -550 - 5376.51 \approx -5926.51 \text{ J/mol·K} \] **Step 4: Final Result:** The total entropy change (ΔS_total) for the reaction is approximately: \[ \Delta S_{\text{total}} \approx -5926.51 \text{ J/mol·K} \]
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