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Current Question (ID: 18501)

Question:

$\text{Ellingham's diagram for the formation of } \text{CO}_2 \text{ is a straight line in the given graph. This is due to:}$ $\begin{array}{c} c(s) + \text{o}_2(g) \rightarrow \text{CO}_2(g) \\ \Delta G^\circ \\ T(K) \end{array}$

Options:

1. $\text{Increase in entropy during } \text{CO}_2 \text{ formation.}$
2. $\text{Decrease in entropy during } \text{CO}_2 \text{ formation.}$
3. $\text{Entropy remains constant during } \text{CO}_2 \text{ formation.}$
4. $\text{Can not be predicted.}$

Solution:

$\text{HINT: Entropy remains constant during } \text{CO}_2 \text{ formation.}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{An Ellingham diagram is a graph showing the temperature dependence of the stability of compounds. This analysis is usually used to evaluate the ease of reduction of metal oxides and sulfides.}$ $\text{In metallurgy, the Ellingham diagram is used to predict the equilibrium temperature between a metal, its oxide, and oxygen — and by extension, reactions of metal with sulfur, nitrogen, and other non-metals.}$ $\text{The diagrams are useful in predicting the conditions under which ore will be reduced to its metal. The analysis is thermodynamic in nature and ignores reaction kinetics.}$ $\text{Thus, processes that are predicted to be favorable by the Ellingham diagram can still be slow.}$ $\text{STEP 2:}$ $\text{Here } \Delta S = 0$ $\therefore \Delta G = \Delta H - T \Delta S$ $\Delta G \text{ is constant with temperature.}$ $\text{This shows that Entropy remains constant during } \text{CO}_2 \text{ formation.}$

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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
}