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Current Question (ID: 21034)

Question:
$\text{For a particular reversible reaction at temperature } T, \Delta H \text{ and } \Delta S \text{ were found to be both } +\text{ve. If } T_e \text{ is the temperature at equilibrium, the reaction would be spontaneous when:}$ $1. \ T = T_e$ $2. \ T_e > T$ $3. \ T > T_e$ $4. \ T_e \text{ is 5 times } T$
Options:
  • 1. $T = T_e$
  • 2. $T_e > T$
  • 3. $T > T_e$
  • 4. $T_e \text{ is 5 times } T$
Solution:
$\text{Hint: Use the formula, that is, } \Delta G = \Delta H - T \Delta S$ $\text{For a spontaneous reaction, } \Delta G \text{ value must be negative.}$ $\text{The formula of } \Delta G \text{ is as follows:}$ $\Delta G = \Delta H - T \Delta S$ $\text{It is given that, } \Delta H \text{ and } \Delta S \text{ value is positive.}$ $\text{At equilibrium, } \Delta G \text{ value is zero.}$ $0 = \Delta H - T_e \Delta S$ $\text{The value of } T \text{ must be greater than the equilibrium temperature. So, that value of } \Delta G \text{ must be negative.}$

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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
}