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Current Question (ID: 21047)

Question:
$\text{If a reaction is non-spontaneous at the freezing point of water but is spontaneous at the boiling point of water, then:}$
Options:
  • 1. $\Delta H \ +ve \ \Delta S \ +ve$
  • 2. $\Delta H \ -ve \ \Delta S \ -ve$
  • 3. $\Delta H \ -ve \ \Delta S \ +ve$
  • 4. $\Delta H \ +ve \ \Delta S \ -ve$
Solution:
$\text{Hint: } \Delta G = -ve \text{ means spontaneous reaction.}$ $\text{For spontaneity means, } \Delta G = -ve$ $\Delta G = \Delta H - T \Delta S$ $\Delta H, \text{ for endothermic process } +ve$ $\text{Entropy decreases as temperature decreases. At lower temperature, } \Delta S \text{ is low hence, overall value of } \Delta G = +ve.$ $\text{But at high temperature } T \text{ value and } \Delta S \text{ value both increases. } T \Delta S \text{ became more negative and will be greater than } \Delta H, \text{ hence } \Delta G = -ve, \text{ spontaneous}$

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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
}