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Current Question (ID: 8301)

Question:

$\text{The enthalpy of sublimation of a substance is equal to:}$

Options:

1. $\text{Enthalpy of fusion} + \text{Enthalpy of vaporization}$
2. $\text{Enthalpy of fusion}$
3. $\text{Enthalpy of vaporization}$
4. $\text{Twice the enthalpy of vaporization}$

Solution:

$\text{Hint: Sublimation reaction in which solid substance directly convert into gas formed.}$ $\text{Step 1:}$ $\text{Sublimation process is a direct conversion of solid to vapour; solid} \rightarrow \text{vapour}$ $\text{Standard enthalpy of sublimation is the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard pressure (1bar).}$ $\text{Step 2:}$ $\text{The sublimation process occurs in two steps, we have solid} \rightarrow \text{liquid} \rightarrow \text{vapour}$ $\text{solid} \rightarrow \text{liquid requires enthalpy of fusion and liquid} \rightarrow \text{vapour requires enthalpy of vaporization.}$ $\text{Hence, Enthalpy of sublimation of a substance is equal to Enthalpy of fusion + Enthalpy of vaporization}$

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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
}