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Current Question (ID: 21073)

Question:
$\text{If the equilibrium constant for } A \rightleftharpoons B + C \text{ is } K_{\text{eq}}^{(1)}$ $\text{and that of } B + C \rightleftharpoons P \text{ is } K_{\text{eq}}^{(2)}, \text{ the equilibrium constant for } A \rightleftharpoons P \text{ is:}$
Options:
  • 1. $K_{\text{eq}}^{(2)} - K_{\text{eq}}^{(1)}$
  • 2. $K_{\text{eq}}^{(1)} \times K_{\text{eq}}^{(2)}$
  • 3. $K_{\text{eq}}^{(1)} + K_{\text{eq}}^{(2)}$
  • 4. $K_{\text{eq}}^{(2)} / K_{\text{eq}}^{(1)}$
Solution:
$\text{Hint: Multiply equation 1 and 2 equilibrium constant}$ $A \rightleftharpoons B + C \quad K_{\text{eq}}^{(1)} \quad (1)$ $B + C \rightleftharpoons P \quad K_{\text{eq}}^{(2)} \quad (2)$ $\text{Step 2:}$ $\text{Add the two equations so that final equation is obtained. If two equations are added and final equal is obtained then equilibrium constant of two equations must be multiple, so that equilibrium constant of final equation is obtained.}$ $\text{Add equation 1 and 2 as follows:}$ $\text{for } A \rightleftharpoons P \quad K_{\text{equ}}$ $K_{\text{equ}} = K_{\text{eq}}^{(1)} \times K_{\text{eq}}^{(2)}$

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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
}