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Question:
$\text{H}^+_{(aq)} + \text{OH}^-_{(aq)} \rightarrow \text{H}_2\text{O}_{(l)} \quad \text{S}^{\circ}(298 \text{ K})$ $\quad -10.7 \text{ JK}^{-1} \quad\quad +70 \text{ JK}^{-1}\text{ mol}^{-1}$ $\text{Standard entropy change for the above reaction is:}$
Options:
  • 1. $60.3 \text{ JK}^{-1} \text{ mol}^{-1}$
  • 2. $80.7 \text{ JK}^{-1} \text{ mol}^{-1}$
  • 3. $-70 \text{ JK}^{-1} \text{ mol}^{-1}$
  • 4. $+10.7 \text{ JK}^{-1} \text{ mol}^{-1}$
Solution:
$\text{Hint: Use equation,}$ $\Delta \text{S} = \Delta \text{S}_\text{P} - \Delta \text{S}_\text{R}$ $\text{To calculate the standard entropy change}$ $(\Delta \text{S}^{\circ})$ $\text{for the reaction, we need to find the difference between the entropy of products and the entropy of reactants:}$ $\Delta \text{S}^{\circ} = \Delta \text{S}_\text{P} - \Delta \text{S}_\text{R}$ $\text{For this reaction:}$ $\text{H}^+_{(aq)} + \text{OH}^-_{(aq)} \rightarrow \text{H}_2\text{O}_{(l)}$ $\Delta \text{S}^{\circ} = \text{S}^{\circ}_{\text{H}_2\text{O}} - (\text{S}^{\circ}_{\text{H}^+} + \text{S}^{\circ}_{\text{OH}^-})$ $\text{From the given data:}$ $\text{S}^{\circ}_{\text{H}^+} = \text{not explicitly given but included in calculation}$ $\text{S}^{\circ}_{\text{OH}^-} = -10.7 \text{ JK}^{-1}$ $\text{S}^{\circ}_{\text{H}_2\text{O}} = +70 \text{ JK}^{-1} \text{ mol}^{-1}$ $\text{We can simplify the calculation as shown in the hint:}$ $\Delta \text{S}^{\circ} = \text{S}^{\circ}_{\text{H}_2\text{O}} - \text{S}^{\circ}_{\text{OH}^-}$ $\Delta \text{S}^{\circ} = 70 - (-10.7)$ $\Delta \text{S}^{\circ} = 70 + 10.7$ $\Delta \text{S}^{\circ} = 80.7 \text{ JK}^{-1} \text{ mol}^{-1}$ $\text{Therefore, the standard entropy change for the reaction is}$ $80.7 \text{ JK}^{-1} \text{ mol}^{-1}$, $\text{which corresponds to option 2.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}