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Current Question (ID: 21043)

Question:
$\text{If at 298 K the bond energies of C-H, C-C, C = C and H-H bonds are respectively 414, 347, 615, and 435 kJ mol}^{-1}\text{, the value of enthalpy change for the reaction at 298 K will be:}$ $\text{H}_2\text{C} = \text{CH}_2(g) + \text{H}_2(g) \rightarrow \text{H}_3\text{C} - \text{CH}_3(g)$
Options:
  • 1. $+250 \text{ kJ}$
  • 2. $-250 \text{ kJ}$
  • 3. $+125 \text{ kJ}$
  • 4. $-125 \text{ kJ}$
Solution:
$\text{Hint: Use the formula, (Bond energy)}_{\text{reactants}} - (\text{BE})_{\text{products}}$ $\text{Calculate the value of enthalpy change for the reaction is as follows:}$ $\text{CH}_2 = \text{CH}_2 + \text{H}_2 \rightarrow \text{CH}_3 - \text{CH}_3$ $\Delta H = (\text{BE})_{\text{reactants}} - (\text{BE})_{\text{products}}$ $= 4(\text{BE})_{\text{C-H}} + (\text{BE})_{\text{C=C}} + (\text{BE})_{\text{H-H}}$ $- [6(\text{BE})_{\text{C-H}} + (\text{BE})_{\text{C-C}}]$ $= -125 \text{ kJ}$

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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
}