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Question:
$\text{The enthalpy of formation of CO}_2\text{(g), H}_2\text{O(l) and propene (g) are } -393.5\text{, } -285.8\text{, and } 20.42 \text{ kJ mol}^{-1} \text{ respectively. The enthalpy change for the combustion of cyclopropane at 298 K will be:}$ $(\text{The enthalpy of isomerization of cyclopropane to propene is } -33.0 \text{ kJ mol}^{-1}\text{. })$
Options:
  • 1. $-1021.32 \text{ kJ mol}^{-1}$
  • 2. $-2091.32 \text{ kJ mol}^{-1}$
  • 3. $-5021.32 \text{ kJ mol}^{-1}$
  • 4. $-3141.32 \text{ kJ mol}^{-1}$
Solution:
$\text{Hint: Hess's law of constant heat summation}$ $\text{Step 1: Identify the relevant reactions and given data}$ $\text{We need to find the enthalpy change for the combustion of cyclopropane (C}_3\text{H}_6\text{). The given data includes:}$ $\Delta\text{H}_f(\text{CO}_2) = -393.5 \text{ kJ mol}^{-1}$ $\Delta\text{H}_f(\text{H}_2\text{O}) = -285.8 \text{ kJ mol}^{-1}$ $\Delta\text{H}_f(\text{propene}) = 20.42 \text{ kJ mol}^{-1}$ $\text{The enthalpy of isomerization of cyclopropane to propene} = -33.0 \text{ kJ mol}^{-1}$ $\text{Step 2: Write out the combustion reaction for propene (C}_3\text{H}_6\text{)}$ $\text{C}_3\text{H}_6\text{(g)} + \frac{9}{2}\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}$ $\text{Step 3: Calculate the enthalpy change for the combustion of propene using Hess's law}$ $\Delta\text{H}_{\text{combustion(propene)}} = [3 \times \Delta\text{H}_f(\text{CO}_2) + 3 \times \Delta\text{H}_f(\text{H}_2\text{O})] - [\Delta\text{H}_f(\text{propene}) + \frac{9}{2} \times \Delta\text{H}_f(\text{O}_2)]$ $\text{Since } \Delta\text{H}_f(\text{O}_2) = 0 \text{ (as it's an element in its standard state):}$ $\Delta\text{H}_{\text{combustion(propene)}} = [3 \times (-393.5) + 3 \times (-285.8)] - [20.42]$ $= [-1180.5 - 857.4] - [20.42]$ $= -2037.9 - 20.42$ $= -2058.32 \text{ kJ mol}^{-1}$ $\text{Step 4: Use the enthalpy of isomerization to find the enthalpy change for the combustion of cyclopropane}$ $\text{Given that the enthalpy of isomerization of cyclopropane to propene is } -33.0 \text{ kJ mol}^{-1}\text{:}$ $\text{cyclopropane(g)} \rightarrow \text{propene(g)}; \Delta\text{H} = -33.0 \text{ kJ mol}^{-1}$ $\text{Using Hess's law, we can combine this with the combustion of propene:}$ $\Delta\text{H}_{\text{combustion(cyclopropane)}} = \Delta\text{H}_{\text{isomerization}} + \Delta\text{H}_{\text{combustion(propene)}}$ $= -33.0 \text{ kJ mol}^{-1} + (-2058.32 \text{ kJ mol}^{-1})$ $= -2091.32 \text{ kJ mol}^{-1}$ $\text{Therefore, the enthalpy change for the combustion of cyclopropane at 298 K is } -2091.32 \text{ kJ mol}^{-1}$

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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
}