Import Question JSON

Current Question (ID: 8419)

Question:

$\text{At } 450 \text{ K, } K_p = 2.0 \times 10^{10} \text{ bar}^{-1} \text{ for the given reaction at equilibrium}$ $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$ $\text{The value of } K_c \text{ at this temperature would be:}$

Options:

1. $7.48 \times 10^{12} \text{ M}^{-1}$
2. $6.56 \times 10^{11} \text{ M}^{-1}$
3. $7.48 \times 10^{11} \text{ M}^{-1}$
4. $1.23 \times 10^{10} \text{ M}^{-1}$

Solution:

$\text{Hint: Use } K_p = K_c((RT))^{\Delta n_g}$ $\text{Step 1: Write down the given information}$ $T = 450 \text{ K and } K_p = 2 \times 10^{10}/\text{bar}$ $\text{Reaction: } 2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)};$ $\Delta n_g = 2 - (2 + 1) = -1$ $\text{Step 2: Calculate } K_c$ $\text{We know that: } K_p = K_c((RT))^{\Delta n_g}$ $\text{Thus, } K_c = \frac{K_p}{((RT))^{\Delta n_g}} = \frac{2.0 \times 10^{10}(\text{bar})^{-1}}{[(0.083 \text{ L bar K}^{-1}(\text{mol})^{-1})(450\text{K})]^{-1}}$ $K_c = 2.0 \times 10^{10} \times 0.083 \times 450 \text{ L(mol)}^{-1}$ $K_c = 7.48 \times 10^{11} \text{ M}^{-1}$

Import JSON File

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
}