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Current Question (ID: 8333)

Question:

$\text{For the reaction,}$ $2\text{Cl}(g) \rightarrow \text{Cl}_2(g)$, $\text{the value of}$ $\Delta\text{H}$ $\text{and}$ $\Delta\text{S}$ $\text{are respectively:}$

Options:

1. $\Delta\text{H} = 0$, $\Delta\text{S} = -\text{ve}$
2. $\Delta\text{H} = 0$, $\Delta\text{S} = 0$
3. $\Delta\text{H} = -\text{ve}$, $\Delta\text{S} = -\text{ve}$
4. $\Delta\text{H} = +\text{ve}$, $\Delta\text{S} = +\text{ve}$

Solution:

$\text{Hint: If number of gaseous particles in products is decreasing, entropy will also decrease.}$ $\text{Explanation:}$ $\text{For the reaction : }$ $2\text{Cl}(g) \rightarrow \text{Cl}_2(g)$ $\text{Bond formation is taking place, i.e. energy is being released. Hence,}$ $\Delta\text{H is -ve.}$ $\text{Also, two moles of atoms have more randomness than one mole of a molecule. Since randomness is decreased,}$ $\Delta\text{S is -ve for the given reaction.}$ $\text{In this reaction:}$ $\text{1. Bond formation occurs, releasing energy, so}$ $\Delta\text{H}$ $\text{is negative}$ $\text{2. The number of gaseous particles decreases from 2 to 1, reducing randomness, so}$ $\Delta\text{S}$ $\text{is negative}$ $\text{Therefore, the correct option is 3:}$ $\Delta\text{H} = -\text{ve}$, $\Delta\text{S} = -\text{ve}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}