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Current Question (ID: 21033)

Question:

$\text{Consider the reaction:}$ $4\text{NO}_2(g) + \text{O}_2(g) \rightarrow 2\text{N}_2\text{O}_5(g) \quad \Delta_rH = -111 \text{ kJ}$ $\text{If } \text{N}_2\text{O}_5(s) \text{ is formed instead of } \text{N}_2\text{O}_5(g) \text{ in the above reaction, the } \Delta_rH \text{ value will be:}$ $\text{(Given: } \Delta H \text{ of sublimation for } \text{N}_2\text{O}_5 \text{ is } 54 \text{ kJ mol}^{-1})$

Options:

1. $+54 \text{ kJ}$
2. $+219 \text{ kJ}$
3. $-219 \text{ kJ}$
4. $-165 \text{ kJ}$

Solution:

$\text{Hint: Apply Hess's Law of Constant Heat Summation concept}$ $\text{Step 1:}$ $\text{Consider the following reaction as follows:}$ $4\text{NO}_2(g) + \text{O}_2(g) \rightarrow 2\text{N}_2\text{O}_5(g) \quad \Delta H_1 = -111 \text{ kJ} \quad \ldots(1)$ $2\text{N}_2\text{O}_5(s) \rightarrow 2\text{N}_2\text{O}_5(g) \quad \Delta H_2 = 108 \text{ kJ} \quad \ldots(2)$ $\text{The sublimation energy for 1 mol of } \text{N}_2\text{O}_5 \text{ is } 54 \text{ kJ.}$ $\text{In the reaction, 2 moles of } \text{N}_2\text{O}_5 \text{ is formed as a product.}$ $\text{Hence, enthalpy of sublimation is } 54 \times 2 = 108 \text{ kJ}$ $\text{The desired equation is as follows:}$ $4\text{NO}_2(g) + \text{O}_2(g) \rightarrow \text{N}_2\text{O}_5(s) \quad \Delta H_3 = ? \quad \ldots(3)$ $\text{Step 2:}$ $\text{Reverse equation 2 and add with equation 1, so that, equation 3 is obtained.}$ $\text{Same scenario will be applied for } \Delta H \text{ values.}$ $\Delta H_3 = H_1 + \Delta H_2$ $= -111 + (-108)$ $= -219 \text{ kJ}$

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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
}