Import Question JSON

Current Question (ID: 8380)

Question:

$\text{For the following reaction,}$ $\text{H}_2\text{(g)}+\text{I}_2\text{(g)} \rightleftharpoons \text{2HI(g)} \text{ at } 250^{\circ} \text{C,}$ $\text{The effect on the state of equilibrium on doubling the volume of the system will be:}$

Options:

1. $\text{Shift to the reactant side}$
2. $\text{Shift to the product side}$
3. $\text{No effect on the state of equilibrium}$
4. $\text{Liquefaction of HI}$

Solution:

$\text{Hint: If the number of mole changes then change in volume also affect the equilibrium.}$ $\text{If the number of moles of reactants and products is equal then the change in volume did not affect the equilibrium condition.}$ $\text{The reaction is given below:}$ $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons \text{2HI(g)} \text{ at } 250^{\circ}\text{C,}$ $\text{Here, the total number of mole of reactants and products are equal both the side. Thus, the change in the volume did not affect the state of equilibrium.}$ $\text{According to Le Chatelier's principle, when volume is doubled (pressure is halved), the equilibrium shifts in the direction where the number of moles of gas increases. However, in this reaction, there are 2 moles of gas on the reactant side (H}_2 \text{ and I}_2\text{) and 2 moles of gas on the product side (2HI). Since the number of gaseous moles is the same on both sides, a change in volume or pressure will have no effect on the equilibrium position.}$ $\text{Hence, option 3 is the correct answer.}$

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
}