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Current Question (ID: 21076)

Question:

$\text{For the reaction } \text{A}_{(g)} \rightarrow \text{B}_{(g)}, \text{ the value of the equilibrium constant at } 300 \text{ K and } 1 \text{ atm is equal to } 100.0. \text{ The value of } \Delta_r G \text{ for the reaction at } 300 \text{ K and } 1 \text{ atm in } \text{J mol}^{-1} \text{ is } -xR, \text{ where } x \text{ is:}$ $\text{(R = 8.31 J mol}^{-1} \text{ K}^{-1} \text{ and ln } 10 = 2.3)}$

Options:

1. $1400$
2. $1380$
3. $1360$
4. $1340$

Solution:

$\text{Hint: } \Delta G^\circ = -RT \ln K_p$ $\text{The relation between } \Delta G^\circ \text{ and } K_p \text{ is as follows:}$ $\Delta G^\circ = -RT \ln K_p$ $= -R(300)(2) \ln(10)$ $= -R(300 \times 2 \times 2.3)$ $\Delta G^\circ = -1380R$

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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
}