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Current Question (ID: 21128)

Question:

$\text{The exothermic formation of ClF}_3 \text{ is represented by the equation:}$ $\text{Cl}_2(g) + 3\text{F}_2(g) \rightleftharpoons 2\ \text{ClF}_3(g); \ \Delta H_r = -329 \ \text{kJ}$ $\text{Which of the following will increase the quantity of ClF}_3 \text{ in an equilibrium mixture of Cl}_2, \text{ F}_2, \text{ and ClF}_3?$

Options:

1. $\text{Adding F}_2$
2. $\text{Increasing the volume of the container}$
3. $\text{Removing Cl}_2$
4. $\text{Increasing the temperature}$

Solution:

$\text{Hint: If reaction proceed in forward direction then concentration of ClF}_3 \text{ increases.}$ $1. \text{ The reaction is an example of an exothermic reaction. If the forward reaction is exothermic then the backward reaction is endothermic and on increasing the temperature backward reaction is favored. Thus, ClF}_3 \text{ concentration decreases.}$ $2. \text{ Removing Cl}_2 \text{ decreases the concentration of Cl}_2 \text{ thus, the reaction will proceed in a backward direction so that Cl}_2 \text{ concentration increases. Thus, ClF}_3 \text{ concentration decreases.}$ $3. \text{ The volume is inversely proportional to pressure hence volume increases the pressure is decreases. Hence, reaction will proceed in the direction where number of mole is high. Thus in backward direction.}$ $4. \text{ Adding F}_2 \text{ will increases the concentration of reactant thus reaction will proceed in forward direction. The concentration of ClF}_3 \text{ increases.}$ $\text{Thus, option first is the correct answer.}$

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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
}