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Current Question (ID: 21166)

Question:
$\text{Consider the following equilibrium reaction:}$ $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ $\text{At equilibrium, if the number of molecules of } \text{H}_2, \text{I}_2, \text{ and HI are equal, and the equilibrium constant } K_p = t \times 10^{-1}, \text{ the value of } t \text{ is:}$
Options:
  • 1. 10.0
  • 2. 0.01
  • 3. 0.10
  • 4. 1.0 (Correct)
Solution:
$\text{Hint: } K_p = \frac{(P_{\text{HI}})^2}{(P_{\text{H}_2})(P_{\text{I}_2})}$ $\text{Explanation:}$ $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ $K_p = \frac{(P_{\text{HI}})^2}{(P_{\text{H}_2})(P_{\text{I}_2})} = 1$ $K_p = t \times 10^{-1}$ $t = 10$

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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
}