Import Question JSON

Current Question (ID: 8404)

Question:

\text{Given the reaction: } 2\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \text{A sample of HI(g) is placed in a flask at a pressure of 0.2 atm.} \text{At equilibrium, the partial pressure of HI(g) is 0.04 atm.} \text{The } K_P \text{ for the given equilibrium would be:}

Options:

1. $2.0$
2. $3.5$
3. $4.0$
4. $2.6$

Solution:

$\text{Hint: The formula of an equilibrium constant is } K_P = \frac{P_{\text{H}_2} \times P_{\text{I}_2}}{(P_{\text{HI}})^2}$ $\text{Explanation:}$ $\text{Step 1:}$ $K_P \text{ for the given equilibrium} = 4.0$ $\text{Calculate the partial pressure of products at equilibrium}$ $\text{At equilibrium, the partial pressure of HI is 0.04 atm. Therefore, a decrease in the pressure of HI is } 0.2 - 0.04 = 0.16$ $\text{The reaction is: } 2\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)}$ $\text{Initial pressure} \quad 0.2 \quad\quad\quad 0 \quad\quad\quad 0$ $\text{At equilibrium} \quad 0.04 \quad\quad \frac{0.16}{2} = 0.08 \quad\quad \frac{0.16}{2} = 0.08$ $\text{Step 2:}$ $\text{Calculate the equilibrium constant value is as follows:}$ $K_P = \frac{P_{\text{H}_2} \times P_{\text{I}_2}}{(P_{\text{HI}})^2}$ $= \frac{0.08 \times 0.08}{(0.04)^2}$ $= 4$ $\text{Hence, } K_P \text{ for the given equilibrium is 4.}$

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
}