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Current Question (ID: 8275)

Question:

$\text{For the reaction, } 2\text{N}_2\text{(g)}+\text{O}_2\text{(g)}\rightarrow 2\text{N}_2\text{O(g)}, \text{ at 298K } \Delta\text{H} \text{ is } 164 \text{ kJ mol}^{-1}. \text{ The } \Delta\text{E} \text{ of the reaction is-}$

Options:

1. $166.5 \text{ kJ mol}^{-1}$
2. $141.5 \text{ kJ mol}^{-1}$
3. $104.0 \text{ kJ mol}^{-1}$
4. $-169 \text{ kJ mol}^{-1}$

Solution:

$\text{Hint: } \Delta\text{H} = \Delta\text{E} + \Delta \text{nRT}$ $\text{Step 1:}$ $\text{The relation between } \Delta\text{E} \text{ and } \Delta\text{H} \text{ is}$ $\Delta\text{H} = \Delta\text{E} + \Delta\text{n}_g \text{RT}$ $\text{Calculate the } \Delta\text{n}_g \text{ value for the reaction:}$ $\Delta\text{n}_g = \text{n(product)} - \text{n(reactant)}$ $= 2 - (2+1)$ $= 2 - 3$ $= -1$ $\text{Step 2:}$ $\text{Rearranging the equation to solve for } \Delta\text{E:}$ $\Delta\text{E} = \Delta\text{H} - (\Delta\text{n}_g) \text{RT}$ $\Delta\text{E} = \Delta\text{H} - (-1) \text{RT}$ $\Delta\text{E} = \Delta\text{H} + \text{RT}$ $\text{Now substituting the values:}$ $\Delta\text{E} = 164 + (8.314 \times 10^{-3} \times 298)$ $\Delta\text{E} = 164 + 2.5$ $\Delta\text{E} \approx 166.5 \text{ kJ mol}^{-1}$ $\text{Note: There is a decrease in the number of moles of gas during the reaction, which means } \Delta\text{n}_g \text{ is negative. This results in } \Delta\text{E} \text{ being greater than } \Delta\text{H} \text{ for this reaction.}$

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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
}