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Current Question (ID: 8331)

Question:

$\text{At a temperature of 300K, what is the entropy change for the reaction given below?}$ $2\text{H}_2 \text{ (g)} + \text{O}_2 \text{ (g)} \rightarrow 2\text{H}_2\text{O(l)}$ $\text{Standard entropies of H}_2 \text{ (g), O}_2\text{(g) and H}_2\text{O(l) are 126.6, 201.20 and 68.0 JK}^{-1}\text{mol}^{-1} \text{ respectively.}$

Options:

1. $-318.4 \text{ JK}^{-1}\text{mol}^{-1}$
2. $318.4 \text{ JK}^{-1}\text{mol}^{-1}$
3. $31.84 \text{ JK}^{-1}\text{mol}^{-1}$
4. $\text{None of the above}$

Solution:

$\text{Hint: } \Delta\text{S}_{\text{reaction}} = \Sigma\text{S}_{\text{product}} - \Sigma\text{S}_{\text{reactant}}$ $\text{Step 1: Identify the given information}$ $\text{The reaction is: } 2\text{H}_2 \text{ (g)} + \text{O}_2 \text{ (g)} \rightarrow 2\text{H}_2\text{O(l)}$ $\text{Given standard entropies:}$ $\text{S}^{\circ}\text{(H}_2\text{, g)} = 126.6 \text{ JK}^{-1}\text{mol}^{-1}$ $\text{S}^{\circ}\text{(O}_2\text{, g)} = 201.20 \text{ JK}^{-1}\text{mol}^{-1}$ $\text{S}^{\circ}\text{(H}_2\text{O, l)} = 68.0 \text{ JK}^{-1}\text{mol}^{-1}$ $\text{Step 2: Calculate the entropy change using the formula}$ $\text{The entropy change for a reaction is given by:}$ $\Delta\text{S}_{\text{reaction}} = \Sigma\text{S}_{\text{products}} - \Sigma\text{S}_{\text{reactants}}$ $\text{For this reaction:}$ $\Delta\text{S}_{\text{reaction}} = \Sigma\text{S}_{\text{products}} - \Sigma\text{S}_{\text{reactants}}$ $= 2 \times \text{S}^{\circ}\text{(H}_2\text{O, l)} - [2 \times \text{S}^{\circ}\text{(H}_2\text{, g)} + \text{S}^{\circ}\text{(O}_2\text{, g)}]$ $\text{Step 3: Substitute the values and calculate}$ $\Delta\text{S}_{\text{reaction}} = 2 \times 68.0 \text{ JK}^{-1}\text{mol}^{-1} - [2 \times 126.6 \text{ JK}^{-1}\text{mol}^{-1} + 201.20 \text{ JK}^{-1}\text{mol}^{-1}]$ $= 136.0 \text{ JK}^{-1}\text{mol}^{-1} - [253.2 \text{ JK}^{-1}\text{mol}^{-1} + 201.20 \text{ JK}^{-1}\text{mol}^{-1}]$ $= 136.0 \text{ JK}^{-1}\text{mol}^{-1} - 454.4 \text{ JK}^{-1}\text{mol}^{-1}$ $= -318.4 \text{ JK}^{-1}\text{mol}^{-1}$ $\text{Therefore, the entropy change for the reaction is } -318.4 \text{ JK}^{-1}\text{mol}^{-1}$ $\text{The negative value of entropy change makes physical sense because in this reaction, three moles of gaseous reactants (2 moles of H}_2\text{ and 1 mole of O}_2\text{) are converted to two moles of liquid water. When gases are converted to liquids, there is a significant decrease in disorder or randomness, as molecules in the liquid state have much less freedom of movement compared to the gas state. This decrease in disorder corresponds to a negative entropy change.}$

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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
}