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Current Question (ID: 8405)

Question:

\text{For the reaction, } 2\text{NOCl(g)} \rightleftharpoons 2\text{NO(g)} + \text{Cl}_2\text{(g)}; K_p = 1.8 \times 10^{-2} \text{ atm at } 500 \text{ K.} \text{The value of } K_c \text{ for above mentioned reaction would be:}

Options:

1. $4.33 \times 10^{-4} \text{ mol L}^{-1}$
2. $4.33 \times 10^{4} \text{ mol L}^{-1}$
3. $1.65 \times 10^{-5} \text{ mol L}^{-1}$
4. $2.39 \times 10^{-3} \text{ mol L}^{-1}$

Solution:

\text{Hint: Use } K_p \text{ and } K_c \text{ relationship.} \text{Step 1: Find out the value of } \Delta n_g \text{ for the given reaction.} \text{Given reaction: } 2\text{NOCl(g)} \rightleftharpoons 2\text{NO(g)} + \text{Cl}_2\text{(g)} \text{Here,} \Delta n_g = 3 - 2 = 1; \quad K_p = 1.8 \times 10^{-2} \text{Step 2: Use relationship between } K_p \text{ and } K_c \text{We know that } K_p = K_c(RT)^{\Delta n_g} 1.8 \times 10^{-2} = K_c(0.0821 \times 500) K_c = \frac{1.8 \times 10^{-2}}{0.0821 \times 500} = 4.33 \times 10^{-4} \text{ mol L}^{-1}

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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
}