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Current Question (ID: 8366)

Question:

$\text{Equilibrium is represented by:}$

Options:

1. $\Delta H = 0$
2. $\Delta G_{\text{Total}} = 0$
3. $\Delta S_{\text{Total}} = 0$
4. $\Delta E = 0$

Solution:

$\text{Hint: At the equilibrium, } \Delta G_{\text{system}} \text{ and } \Delta S_{\text{total}} \text{ will be zero.}$\n\n$\text{Explanation:}$\n\n$\text{At equilibrium, rate of forward reaction is equal to rate of backward reaction. The formula of Gibbs free energy is as follows:}$\n\n$\Delta G = \Delta H - T\Delta S$\n\n$\text{For a spontaneous reaction, } \Delta G \text{ must be negative and } \Delta S \text{ must be positive. At equilibrium, } \Delta G_{\text{system}} \text{ and } \Delta S_{\text{total}} \text{ will have zero value.}$

Import JSON File

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
}