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Current Question (ID: 21136)

Question:

$\text{For the reaction equilibrium, } \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g), \text{ the concentrations of } \text{N}_2\text{O}_4 \text{ and } \text{NO}_2 \text{ at equilibrium are } 4.8 \times 10^{-2} \text{ and } 1.2 \times 10^{-2} \text{ mol L}^{-1} \text{ respectively. The value of } K_c \text{ for the reaction is:}$

Options:

1. $3.3 \times 10^{2} \text{ mol L}^{-1}$
2. $3 \times 10^{-1} \text{ mol L}^{-1}$
3. $3 \times 10^{-3} \text{ mol L}^{-1}$
4. $3 \times 10^{3} \text{ mol L}^{-1}$

Solution:

$\text{Hint: Use the formula, that is, } K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}$ $\text{Calculate the value of } K_c \text{ as follows:}$ $K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{(1.2 \times 10^{-2})^2}{4.8 \times 10^{-2}} = 3 \times 10^{-3} \text{ mol L}^{-1}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}