Import Question JSON

Current Question (ID: 8344)

Question:

$\text{For the conversion }\text{C}_{(\text{graphite})} \rightarrow \text{C}_{(\text{diamond})}\text{ the }\Delta S\text{ is:}$

Options:

1. $\text{Zero}$
2. $\text{Positive}$
3. $\text{Negative}$
4. $\text{Unknown}$

Solution:

$\text{Hint: Conversion of graphite into diamond is an endothermic reaction}$ $\text{To determine the entropy change (}\Delta S\text{) for the conversion of graphite to diamond, we need to analyze the structural changes and organization in this transformation:}$ $\text{1. Structure of graphite:}$ $\text{- Graphite has a layered structure with loosely bound sheets of carbon atoms}$ $\text{- Carbon atoms are arranged in a planar hexagonal pattern within each layer}$ $\text{- The layers can slide over each other, giving graphite its lubricating properties}$ $\text{- This structure has higher disorder/randomness compared to diamond}$ $\text{2. Structure of diamond:}$ $\text{- Diamond has a rigid three-dimensional tetrahedral crystal structure}$ $\text{- Carbon atoms are held together by strong covalent bonds in all directions}$ $\text{- The structure is extremely ordered and compact}$ $\text{- This structure has lower disorder/randomness compared to graphite}$ $\text{When graphite is converted to diamond, the carbon atoms move from a relatively less ordered structure to a more ordered, rigid structure. Since entropy is a measure of disorder or randomness in a system, this transition results in a decrease in entropy.}$ $\text{Therefore, the entropy change (}\Delta S\text{) for the conversion of graphite to diamond is negative.}$ $\text{This is consistent with the information provided:}$ $\text{- The conversion is endothermic (}\Delta H = +1.9 \text{ kJ/mol}\text{), meaning energy is absorbed}$ $\text{- Diamond has higher internal energy than graphite}$ $\text{- Graphite is more thermodynamically stable than diamond at 25}^{\circ}\text{C}$ $\text{For a process to be spontaneous, the Gibbs free energy change must be negative (}\Delta G < 0\text{). Since:}$ $\Delta G = \Delta H - T\Delta S$ $\text{With positive }\Delta H\text{ and negative }\Delta S\text{, the conversion of graphite to diamond is non-spontaneous at standard conditions, which explains why graphite is the more thermodynamically stable form of carbon at room temperature and atmospheric pressure.}$ $\text{The correct answer is option 3: Negative.}$

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
}