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Current Question (ID: 21105)

Question:

$\text{The equilibrium constant at 298 K for a reaction A + B} \rightleftharpoons \text{C + D is 100.}$ $\text{If the initial concentration of all the four species were 1 M each,}$ $\text{then equilibrium concentration of D (in mol L}^{-1}\text{) will be:}$

Options:

1. $0.182$
2. $0.818$
3. $1.818$
4. $1.182$

Solution:

$\text{Hint: Equilibrium constant formula is } K_{\text{equ}} = \frac{[C] \times [D]}{[B] \times [A]}$ $\text{Step 1:}$ $\text{The reaction is as follows:}$ $\text{A + B} \rightleftharpoons \text{C + D}$ $\text{At } t = 0 \quad 1 \quad 1 \quad 1 \quad 1$ $\text{At equilibrium: } 1-x \quad 1-x \quad 1+x \quad 1+x$ $\text{The expression of the equilibrium constant is as follows:}$ $K_{\text{equ}} = \frac{(x+1) \times (x+1)}{(1-x) \times (1-x)}$ $\text{Step 2:}$ $100 = \frac{(x+1)^2}{(1-x)^2}$ $10 = \frac{x+1}{1-x}$ $10 - 10x = x + 1$ $11x = 9$ $x = \frac{9}{11}$ $x = 0.818 \text{ M}$ $\text{The equilibrium concentration of D is } (1+x).$ $[D] = 1 + x; = 1 + 0.818; = 1.818 \text{ M}$

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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
}