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Current Question (ID: 8397)

Question:
\text{For the following equilibrium, } K_c = 6.3 \times 10^{14} \text{ at 1000 K} \text{NO(g) + O}_3\text{(g) } \to \text{ NO}_2\text{(g) + O}_2\text{(g)} \text{The value of } K_c \text{ for the reverse reaction is:}
Options:
  • 1. $2.33 \times 10^{-16}$
  • 2. $1.59 \times 10^{-15}$ (Correct)
  • 3. $2.67 \times 10^{-13}$
  • 4. $4.47 \times 10^{14}$
Solution:
$\text{Hint: Use properties of the equilibrium constant}$ $\text{Step 1: Write down the given data.}$ $\text{It is given that } K_c \text{ for the forward reaction is } 6.3 \times 10^{14}.$ $\text{Step 2: } K_c \text{ for the reverse reaction will depend on the given } K_c \text{ value}$ $\text{Then, } K'_c \text{ for the reverse reaction will be,}$ $K'_c = \frac{1}{K_c}$ $= \frac{1}{6.3 \times 10^{14}}$ $= 1.59 \times 10^{-15}$

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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
}