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Current Question (ID: 21037)

Question:

$\text{Standard entropy of } \text{X}_2, \text{Y}_2 \text{ and } \text{XY}_3 \text{ are } 60, 40 \text{ and } 50 \text{ J K}^{-1} \text{ mol}^{-1}, \text{ respectively. For the reaction, }$ $\frac{1}{2} \text{X}_2 + \frac{3}{2} \text{Y}_2 \rightarrow \text{XY}_3 \Delta \text{H} = -30 \text{ kJ, to be at equilibrium, the temperature will be}$

Options:

1. $500 \text{ K}$
2. $750 \text{ K}$
3. $1000 \text{ K}$
4. $1250 \text{ K}$

Solution:

$\text{Hint: The formula of entropy is } \Delta S^\circ = \frac{\Delta H^\circ}{T}$ $\text{Step 1:}$ $\text{First, calculate the value of } \Delta S^\circ \text{ as follows:}$ $\text{The given reaction is as follows:}$ $\frac{1}{2} \text{X}_2 + \frac{3}{2} \text{Y}_2 \rightarrow \text{XY}_3$ $\Delta S^\circ = \Delta S^\circ_{\text{XY}_3} - \left( \frac{1}{2} \Delta S^\circ_{\text{X}_2} + \frac{3}{2} \Delta S^\circ_{\text{Y}_2} \right)$ $= 50 - \left( \frac{1}{2} \times 60 + \frac{3}{2} \times 40 \right)$ $= 50 - 90$ $= -40 \text{ J K}^{-1}$ $\text{Step 2:}$ $\text{Calculate the value of temperature as follows:}$ $\Delta S^\circ = \frac{\Delta H^\circ}{T}$ $-40 = \frac{-30 \times 10^3}{T}$ $T = \frac{30 \times 10^3}{40}$ $T = 750 \text{ K}$ $\text{Hence, option second is the correct answer.}$

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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
}