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Current Question (ID: 21077)

Question:

$\text{For a reaction;} \ X \ + \ Y \ \rightleftharpoons \ 2Z, \ 1.0 \ \text{mol of} \ X, \ 1.5 \ \text{mol of} \ Y \ \text{and} \ 0.5 \ \text{mol of} \ Z \ \text{were taken in a} \ 1 \ \text{L vessel and allowed to react. At equilibrium, the concentration of} \ Z \ \text{was} \ 1.0 \ \text{mol L}^{-1}. \ \text{The equilibrium constant of the reaction is} \ \frac{x}{15}. \ \text{The value of} \ x \ \text{is:}$

Options:

1. $24$
2. $13$
3. $16$
4. $19$

Solution:

$\text{Hint: The formula of equilibrium constant is,} \ K_{\text{equ}} = \frac{Z^2}{X \times Y}$ $\text{Step 1:}$ $\text{The concentration of X, Y and Z is as follows:}$ $\text{Concentration} = \frac{\text{number of mole}}{\text{volume of solution}}$ $\text{Hence, volume of solution is 1 L, concentration is equal to number of moles.}$ $\text{So concentration of X, Y and Z is equal to its mole.}$ $\begin{array}{ccc} & X + Y = 2Z \\ t = 0 & 1 & 1.5 & 0.5 \\ \text{At eq.} & 1-x & 1.5-x & 0.5+2x \end{array}$ $\text{The equilibrium concentration of Z is 1. Calculate the value of x as follows:}$ $[Z] = 0.5 + 2x$ $1 = 0.5 + 2x$ $x = 0.25 \ \text{M}$ $\text{The concentration of X and Y at equilibrium is as follows:}$ $[X] = 1 - x = 0.75, \ [Y] = 1.5 - x = 1.25$ $\text{Step 2:}$ $K_{\text{equ}} = \frac{Z^2}{X \times Y}$ $K_{\text{eq.}} = \frac{1^2}{\frac{3}{4} \times \frac{5}{4}} = \frac{16}{15}$

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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
}