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Question:
$\Delta_{\text{vap}}\text{H}^{\circ} (\text{CCl}_4) = 30.5 \text{ kJ mol}^{-1}$ $\Delta_f\text{H}^{\circ} (\text{CCl}_4) = -135.5 \text{ kJ mol}^{-1}$ $\Delta_a\text{H}^{\circ} (\text{C}) = 715.0 \text{ kJ mol}^{-1}$ $\Delta_a\text{H}^{\circ} (\text{Cl}_2) = 242 \text{ kJ mol}^{-1}$ $\text{The enthalpy change for the reaction}$ $\text{CCl}_4 \text{ (g)} \rightarrow \text{C(g)} + 4 \text{ Cl (g) would be:}$
Options:
  • 1. $326 \text{ kJ mol}^{-1}$
  • 2. $1304 \text{ kJ mol}^{-1}$
  • 3. $-328 \text{ kJ mol}^{-1}$
  • 4. $-1304 \text{ kJ mol}^{-1}$
Solution:
$\text{Hint: Hess's Law of Constant Heat Summation}$ $\text{Step 1: Write down the given data}$ $(i)\text{ CCl}_4\text{(l)} \rightarrow \text{CCl}_4\text{(g)}; \Delta_{\text{vap}} \text{H}^{\circ} (\text{CCl}_4) = 30.5 \text{ kJ mol}^{-1}$ $(ii)\text{ C(s)} + 2\text{Cl}_2\text{(g)} \rightarrow \text{CCl}_4\text{(l)}; \Delta_f\text{H}^{\circ} (\text{CCl}_4) = -135.5 \text{ kJ mol}^{-1}$ $(iii)\text{ C(s)} \rightarrow \text{C(g)}; \Delta_a\text{H}^{\circ} (\text{C}) = 715.0 \text{ kJ mol}^{-1}$ $(iv)\text{ Cl}_2\text{(g)} \rightarrow 2\text{Cl(g)}; \Delta_a\text{H}^{\circ} (\text{Cl}_2) = 242 \text{ kJ mol}^{-1}$ $\text{Step 2: Calculate the enthalpy change for required reaction}$ $\text{Reaction: CCl}_4 \text{(g)} \rightarrow \text{C(g)} + 4 \text{ Cl (g)}$ $\text{The enthalpy change for the above reaction can be calculated using the following algebraic calculations:}$ $\text{Equation (iii)} + 2 \times \text{Equation (iv)} - \text{Equation (i)} - \text{Equation (ii)}$ $\Delta\text{H} = \Delta_a\text{H}^{\circ} (\text{C}) + 2[\Delta_a\text{H}^{\circ} (\text{Cl}_2)] - \Delta_{\text{vap}}\text{H}^{\circ} (\text{CCl}_4) - \Delta_f \text{H}^{\circ} (\text{CCl}_4)$ $= 715.0 \text{ kJ mol}^{-1} + 2(242 \text{ kJ mol}^{-1}) - (30.5 \text{ kJ mol}^{-1}) - (-135.5 \text{ kJ mol}^{-1})$ $= 715.0 \text{ kJ mol}^{-1} + 484 \text{ kJ mol}^{-1} - 30.5 \text{ kJ mol}^{-1} + 135.5 \text{ kJ mol}^{-1}$ $= 1304 \text{ kJ mol}^{-1}$ $\text{The positive value indicates that this is an endothermic reaction, requiring energy input.}$ $\text{Therefore, the enthalpy change for the reaction CCl}_4 \text{(g)} \rightarrow \text{C(g)} + 4 \text{ Cl (g) is } 1304 \text{ kJ mol}^{-1}$

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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
}