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Current Question (ID: 8398)

Question:

\text{The value of } \Delta n \text{ for the following reaction will be:} \text{NH}_4\text{Cl}_{(\text{s})} \rightleftharpoons \text{NH}_3_{(\text{g})} + \text{HCl}_{(\text{g})}

Options:

1. $1$
2. $0.5$
3. $1.5$
4. $2$ (Correct)

Solution:

$\text{Hint: } \Delta n_g = \text{number of gaseous molecules of products - number of gaseous molecules of reactants.}$ $\text{Step 1:}$ $\text{The relationship between } K_c \text{ and } K_p \text{ is}$ $K_p = K_c(RT)^{\Delta n}$ $\text{where, } \Delta n = (\text{number of moles of gaseous products})-(\text{number of moles of gaseous reactants})$ $\text{Step 2:}$ $\text{For solid-state reactants or product number of moles is constant and we did not include it.}$ $\text{For the reaction,}$ $\text{NH}_4\text{Cl}_{\text{(s)}} \rightleftharpoons \text{NH}_3\text{(g)} + \text{HI}_{\text{(g)}}$ $\Delta n = 2 - 0 = 2$ $\text{For the given reaction:}$ $\text{NH}_4\text{Cl}_{\text{(s)}} \rightleftharpoons \text{NH}_3\text{(g)} + \text{HCl}_{\text{(g)}}$ $\text{Number of moles of gaseous products = } \text{NH}_3\text{(g)} + \text{HCl}_{\text{(g)} = 1 + 1 = 2$ $\text{Number of moles of gaseous reactants = 0 (since NH}_4\text{Cl is solid)}$ $\text{Therefore, } \Delta n = 2 - 0 = 2$

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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
}