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Current Question (ID: 21172)

Question:
$\text{Consider the following equilibrium reactions along with their respective equilibrium constants:}$ $x \rightleftharpoons y; \; k_1 = 1$ $y \rightleftharpoons z; \; k_2 = 2$ $z \rightleftharpoons w; \; k_3 = 4$ $\text{Now, calculate the overall equilibrium constant } k_{eq} \text{ for the reaction:}$ $x \rightleftharpoons w$ $\text{Choose the correct answer:}$
Options:
  • 1. $10$
  • 2. $8$
  • 3. $2$
  • 4. $4$
Solution:
$x \rightleftharpoons y; \; k_1 = 1 \ldots (i)$ $y \rightleftharpoons z; \; k_2 = 2 \ldots (ii)$ $z \rightleftharpoons w; \; k_3 = 4 \ldots (iii)$ $\text{On adding equation (i), (ii) and (iii)}$ $x \rightleftharpoons w$ $k_{eq} = k_1 \times k_2 \times k_3$ $= 1 \times 2 \times 4 = 8$

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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
}