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Current Question (ID: 8354)

Question:

$\text{The values of } \Delta H \text{ and } \Delta S \text{ for the given reaction are 170 kJ and 170 J K}^{-1}\text{, respectively.}$\n\n$\text{C}_{\text{(graphite)}} + \text{CO}_{2\text{(g)}} \rightarrow 2\text{CO}_{\text{(g)}}$\n\n$\text{This reaction will be spontaneous at:}$

Options:

1. $710 \text{ K}$
2. $910 \text{ K}$
3. $1110 \text{ K}$
4. $510 \text{ K}$

Solution:

$\text{Hint: Use of equation relationship between Gibb's free energy change, enthalpy change, and Temperature.}$\n\n$\text{Step 1:}$\n\n$\text{Given,}$\n$\Delta H = 170 \text{ kJ} = 170 \times 10^3 \text{ J}$\n$\Delta S = 170 \text{ J K}^{-1}$\n\n$\text{Step 2:}$\n\n$\text{Substituting these values in the equation, we can find the value of T.}$\n\n$\Delta G = \Delta H - T \Delta S$\n$\Delta G = 170000 - T(170)$\n\n$\text{Since we know that a reaction is spontaneous if } \Delta G < 0 \text{ and it is only possible if T = 1110 K (option 3rd)}$\n\n$\text{(To solve this question, proceed by putting the T values given in the options one by one and see which value of T will give negative } \Delta G \text{ value for a spontaneous process.)}$

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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
}