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Current Question (ID: 8407)

Question:
\text{The value of the equilibrium constant for a particular reaction is } 1.6 \times 10^{12}. \text{ When the system is in equilibrium, it will include:}
Options:
  • 1. $\text{All reactants}$
  • 2. $\text{Mostly reactants}$
  • 3. $\text{Mostly products}$
  • 4. $\text{Similar amounts of reactants and products}$
Solution:
\text{Hint: The value of equilibrium constant is very large. Then reaction almost completed} \text{Step 1:} \text{If } K_c > 10^3\text{, products predominate over reactants, i.e., if } K_c \text{ is very large, the reaction proceeds nearly to completion. The concentration of the product is very high as compared to the reactant.} \text{Step 2:} A \rightleftharpoons B K_c = \frac{[B]}{[A]} = 1.6 \times 10^{12} \text{The equilibrium constant value is very high, thus, the concentration of the product is more as compared to the reactant.}

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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
}