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Current Question (ID: 18498)

Question:

$\text{The thermodynamic property useful for selecting the reducing agent is:}$

Options:

1. $\text{Gibbs free energy change.}$
2. $\text{Internal energy change.}$
3. $\text{Specific heat capacity.}$
4. $\text{None of the above.}$

Solution:

$\text{HINT: Ellingham Diagram.}$ $\text{The above figure is a plot of Gibbs energy } \Delta G^\circ \text{ vs T for the formation of some oxides.}$ $\text{It can be observed from the above graph that metal can reduce the oxide of other metals, if the standard free energy of formation of the oxide of the former is more negative than the latter.}$ $\text{For example, since } \Delta_f G^\circ (\text{Al} / \text{Al}_2 \text{O}_3) \text{ is more negative than } \Delta_f G^\circ (\text{Cu} / \text{Cu}_2 \text{O}). \text{ Thus, Al can reduce Cu}_2\text{O to Cu, but Cu cannot reduce Al}_2\text{O}_3.$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}