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Current Question (ID: 21167)

Question:

$\text{Consider the following equilibrium reaction:}$ $\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightleftharpoons 2\text{Fe}(s) + 3\text{CO}_2(g)}$ $\text{Which of the following will not affect the equilibrium state?}$ $\text{(I) Addition of Fe}_2\text{O}_3$ $\text{(II) Addition of CO}_2$ $\text{(III) Decreasing the mass of Fe}_2\text{O}_3$ $\text{(IV) Removal of CO}$

Options:

1. $(\text{II}) \text{ and } (\text{IV})$
2. $(\text{I}) \text{ and } (\text{IV})$
3. $(\text{I}) \text{ and } (\text{III})$
4. $\text{All will affect the equilibrium state}$

Solution:

$\text{Hint: Changes in the mass of a pure solid don't affect equilibrium.}$ $\text{As conc. of solid species is constant (active mass = 1) at a given temperature,}$ $\text{for solid addition or decreasing mass do not affect equilibrium state.}$

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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
}