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Current Question (ID: 8409)

Question:

$\text{For the reaction,}$ $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)};$ $K_c = 0.061 \text{ mol}^{-2}\text{L}^2 \text{ at 500K. At a particular instant of time, [N}_2\text{] = 3.0 mol L}^{-1}\text{, [H}_2\text{] = 2.0 mol L}^{-1}\text{ and}$ $\text{[NH}_3\text{] = 0.5 mol L}^{-1}\text{.}$ $\text{True statement among the following is:}$

Options:

1. $\text{Reaction is at equilibrium.}$
2. $\text{Reaction will proceed in the forward direction.}$
3. $\text{Reaction will proceed in the backward direction.}$
4. $\text{Can't predict the direction of the reaction.}$

Solution:

$\text{Hint: If } Q_c = K_c, \text{ the reaction is at equilibrium.}$ $\text{Explanation:}$ $\text{Given reaction:}$ $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ $\text{Conc. at time 't': 3.0, 2.0, 0.5 respectively}$ $\text{Now, we know that}$ $Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$ $Q_c = \frac{(0.5)^2}{(3.0)(2.0)^3} = \frac{0.25}{3.0 \times 8} = \frac{0.25}{24} = 0.0104$ $\text{Given: } K_c = 0.061$ $\text{Since } Q_c < K_c \text{ (0.0104 < 0.061), the reaction will proceed in the forward direction to reach equilibrium.}$

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
}