Import Question JSON

Current Question (ID: 8427)

Question:

\text{A mixture of } 1.57 \text{ mol of } \text{N}_2\text{, } 1.92 \text{ mol of } \text{H}_2\text{, and } 8.13 \text{ mol of } \text{NH}_3 \text{ is introduced into a } 20 \text{ L vessel at } 500 \text{ K.} \text{At this temperature, the equilibrium constant, } K_c \text{ for the reaction} \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \text{is } 1.7 \times 10^2\text{.} \text{The direction of the net reaction is:}

Options:

1. $\text{Reaction is at equilibrium.}$
2. $\text{Reaction will proceed in forwarding direction.}$
3. $\text{Reaction will proceed in the backward direction.}$
4. $\text{Data is not sufficient.}$

Solution:

\text{Hint: If } K_c < Q_c \text{, the reaction will proceed in the backward direction.} \text{The given reaction is} \mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} \rightleftharpoons 2\mathrm{NH}_{3(g)} \text{Given concentration of } [\mathrm{N}_2] = \frac{1.57}{20} \text{ mol L}^{-1}\text{, } [\mathrm{H}_2] = \frac{1.92}{20} \text{ mol L}^{-1} \text{ and } [\mathrm{NH}_3] = \frac{8.13}{20} \text{ mol L}^{-1} \text{Now, the reaction quotient } Q_c Q_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} Q_c = \frac{(0.41)^2}{(0.079)(0.096)^3} Q_c = 2.4 \times 10^3 \text{Since, } Q_c > K_c \text{, the reaction will proceed in the backward direction.}

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
}