Import Question JSON

Current Question (ID: 8418)

Question:

$\text{For the equilibrium of the reaction, } \text{HgO}_{\text{(s)}} \rightleftharpoons \text{Hg}_{\text{(g)}} + \frac{1}{2}\text{O}_2\text{(g)}, K_p \text{ for the reaction at a total pressure of }\text{\"P\"} \text{ will be:}$

Options:

1. $K_P = \frac{2}{3^{3/2}} \text{P}^{3/2}$
2. $K_P = \frac{2}{3^{1/2}} \text{P}^{1/2}$
3. $K_P = \frac{1}{3^{2/3}} \text{P}^{3/2}$
4. $K_P = \frac{1}{3^{2/3}} \text{P}$

Solution:

$\text{Hint: } K_p = P_{\text{Hg}} \times P_{\text{O}_2}^{1/2}$ $\text{Step 1:}$ $\text{HgO}_{\text{(s)}} \rightleftharpoons \text{Hg}_{\text{(g)}} + \frac{1}{2}\text{O}_2\text{(g)}$ $\text{Initial: } 1 \text{ mole HgO, } 0 \text{ Hg, } 0 \text{ O}_2$ $\text{At equilibrium: } (1-x) \text{ HgO, } x \text{ Hg, } \frac{x}{2} \text{ O}_2$ $\text{Solid substance mole remained constant and did not include in equilibrium constant expression.}$ $\text{Total moles at equilibrium } = x + \frac{x}{2} = \frac{3x}{2}$ $\text{Step 2:}$ $\text{Calculate the partial pressure of gaseous products and reactants using Dalton's law of partial pressure}$ $P = P_T \times X(\text{mole fraction})$ $P_{\text{Hg}} = \frac{x}{3x/2} \times P = \frac{2}{3}P$ $P_{\text{O}_2} = \frac{x/2}{3x/2} \times P = \frac{1}{3}P$ $K_p = P_{\text{Hg}} \times P_{\text{O}_2}^{1/2} = \frac{2}{3}P \times \left(\frac{1}{3}P\right)^{1/2} = \frac{2}{3^{3/2}} P^{3/2}$ $\text{Hence, option 1 is the correct answer.}$

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
}