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Current Question (ID: 21036)

Question:

$\text{The oxidizing power of chlorine in an aqueous solution can be determined by the following parameters:}$ $\frac{1}{2} \text{Cl}_2 (g) \xrightarrow{\Delta_{\text{diss}} H^\circ} \text{Cl} (g) \xrightarrow{\Delta_{\text{eg}} H^\circ} \text{Cl}^- (g) \xrightarrow{\Delta_{\text{hyd}} H^\circ} \text{Cl}^- (aq)$ $\text{The energy involved in the conversion of } \frac{1}{2} \text{Cl}_2 (g) \text{ to } \text{Cl}^- (aq) \text{ will be:}$ $\text{Use the following data:}$ $\Delta_{\text{diss}} H^\circ (\text{Cl}_2) = 240 \text{ kJ mol}^{-1}$ $\Delta_{\text{eg}} H^\circ (\text{Cl}) = -349 \text{ kJ mol}^{-1}$ $\Delta_{\text{hyd}} H^\circ (\text{Cl}^-) = -381 \text{ kJ mol}^{-1}$

Options:

1. $-610 \text{ kJ mol}^{-1}$
2. $-850 \text{ kJ mol}^{-1}$
3. $+120 \text{ kJ mol}^{-1}$
4. $+152 \text{ kJ mol}^{-1}$

Solution:

$\text{Hint: } \Delta H^\circ_{\text{reac}} = \frac{1}{2} \Delta H^\circ_{\text{diss}} + \Delta H^\circ_{\text{Eg}} + \Delta H^\circ_{\text{Hyd}}$ $\text{Step 1:}$ $\text{The reaction is as follows:}$ $\frac{1}{2} \text{Cl}_2 (g) \xrightarrow{\Delta_{\text{diss}} H^\circ} \text{Cl} (g) \xrightarrow{\Delta_{\text{eg}} H^\circ} \text{Cl}^- (g) \xrightarrow{\Delta_{\text{hyd}} H^\circ} \text{Cl}^- (aq)$ $\text{Step 2:}$ $\text{Calculate the energy involved in the conversion of } \frac{1}{2} \text{Cl}_2 (g) \text{ to } \text{Cl}^- (aq) \text{ as follows:}$ $\Delta H^\circ_{\text{reac}} = \frac{1}{2} \Delta H^\circ_{\text{diss}} + \Delta H^\circ_{\text{Eg}} + \Delta H^\circ_{\text{Hyd}}$ $\Delta H^\circ_{\text{reac}} = \frac{1}{2} (240) + (-349) + (-381)$ $= -610 \text{ kJ mol}^{-1}$

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{
  "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
}