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Current Question (ID: 8300)

Question:

$\text{The standard enthalpy of combustion at }25^{\circ}C\text{ of hydrogen, cyclohexene }(\text{C}_6\text{H}_{10}),\text{ and cyclohexane }(\text{C}_6\text{H}_{12})\text{ are }-241, -3800\text{ and }-3920\text{ kJ mol}^{-1},\text{ respectively. Calculate the standard enthalpy of hydrogenation of cyclohexene.}$

Options:

1. $-131\text{ kJ mol}^{-1}$
2. $-155\text{ kJ mol}^{-1}$
3. $-167\text{ kJ mol}^{-1}$
4. $-121\text{ kJ mol}^{-1}$

Solution:

$\text{Hint: }\Delta H^{\circ} = \Delta H^{\circ}_{(ii)} + \Delta H^{\circ}_{(i)} - \Delta H^{\circ}_{(iii)}$ $\text{The given data are :}$ $(i)\text{ }\text{H}_2(g)+\frac{1}{2}\text{O}_2(g)\rightarrow \text{H}_2\text{O}(l) ; \quad \Delta H^{\circ} = -241\text{ kJ mol}^{-1}$ $(ii)\text{ }\text{C}_6\text{H}_{10}(g)+\frac{17}{2}\text{O}_2(g)\rightarrow 6\text{CO}_2(g)+5\text{H}_2\text{O}(l)$ $\Delta H^{\circ} = -3800\text{ kJ mol}^{-1}$ $(iii)\text{ }\text{C}_6\text{H}_{12}(g)+9\text{O}_2(g)\rightarrow 6\text{CO}_2(g)+6\text{H}_2\text{O}(l)$ $\Delta H^{\circ} = -3920\text{ kJ mol}^{-1}$ $\text{We have to calculate the enthalpy change for the reaction}$ $\text{C}_6\text{H}_{10}(g)+\text{H}_2(g)\rightarrow \text{C}_6\text{H}_{12}(g)$ $\text{This equation can be obtained by the following manipulations.}$ $\text{Eq.(ii) + Eq.(i) - Eq.(iii)}$ $\text{Carrying out the corresponding manipulations on }\Delta H^{\circ}\text{s, we get}$ $\Delta H^{\circ} = \Delta H^{\circ}_{(ii)} + \Delta H^{\circ}_{(i)} - \Delta H^{\circ}_{(iii)}$ $=(−3800 - 241 + 3920)\text{ kJ mol}^{-1} = -121\text{ kJ mol}^{-1}.$

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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
}