Import Question JSON

Current Question (ID: 8393)

Question:

$K_c = \frac{[\text{NH}_3]^4[\text{O}_2]^5}{[\text{NO}]^4[\text{H}_2\text{O}]^6}$ $\text{The balanced chemical equation corresponding to the above-mentioned expression is:}$

Options:

1. $4\text{NO}_{(g)} + 6\text{H}_2\text{O}_{(g)} \rightleftharpoons 4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)}$
2. $4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)} \rightleftharpoons 4\text{NO}_{(g)} + 6\text{H}_2\text{O}_{(g)}$
3. $2\text{NO}_{(g)} + 3\text{H}_2\text{O}_{(g)} \rightleftharpoons 4\text{NH}_3\text{(g)} + 3\text{O}_2\text{(g)}$
4. $\text{NH}_3\text{(g)} + 3\text{H}_2\text{O}_{(g)} \rightleftharpoons 2\text{NO}_{(g)} + 3\text{O}_2\text{(g)}$

Solution:

$\text{Hint: Use the general expression of } K_c \text{ for a balanced reaction.}$ $\text{Step 1: Write down the given expression of } K_c$ $K_c = \frac{[\text{NH}_3]^4[\text{O}_2]^5}{[\text{NO}]^4[\text{H}_2\text{O}]^6}$ $\text{Step 2: Write down the balanced reaction by seeing } K_c \text{ expression}$ $\text{The balanced chemical equation corresponding to the given expression can be written as:}$ $4\text{NO}_{(g)} + 6\text{H}_2\text{O}_{(g)} \rightleftharpoons 4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)}$ $\text{Analysis of the } K_c \text{ expression:}$ $\text{General form: } K_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}}$ $\text{From the given expression:}$ $\text{• Products (numerator): NH}_3 \text{ with coefficient 4, O}_2 \text{ with coefficient 5}$ $\text{• Reactants (denominator): NO with coefficient 4, H}_2\text{O} \text{ with coefficient 6}$ $\text{Therefore, the balanced equation is:}$ $4\text{NO}_{(g)} + 6\text{H}_2\text{O}_{(g)} \rightleftharpoons 4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)}$ $\text{Verification: The } K_c \text{ expression for this equation would be:}$ $K_c = \frac{[\text{NH}_3]^4[\text{O}_2]^5}{[\text{NO}]^4[\text{H}_2\text{O}]^6}$ ✓

Import JSON File

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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
}