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Current Question (ID: 8365)

Question:

$\text{For the reaction } 2\text{A}_{\text{(g)}} + \text{B}_{\text{(g)}} \rightarrow 2\text{D}_{\text{(g)}}; \Delta U^o = -10.5 \text{ kJ and } \Delta S^o = -44.1 \text{ J K}^{-1}\text{, the value of } \Delta G^o \text{ for the given reaction would be-}$

Options:

1. $1.6 \text{ J}$
2. $-0.16 \text{ kJ}$
3. $0.16 \text{ kJ}$
4. $1.6 \text{ kJ}$

Solution:

$\text{Hint: } \Delta G^o = \Delta H^o - T \Delta S^o$\n\n$\text{Step 1: Calculate } \Delta H^o$\n\n$\text{For the reaction: } 2\text{A}_{\text{(g)}} + \text{B}_{\text{(g)}} \rightarrow 2\text{D}_{\text{(g)}}; \Delta n = 2 - (2 + 1) = -1$\n\n$\text{We know that;}$\n\n$\Delta H^o = \Delta U^o + \Delta n_g RT$\n$\Delta H^o = -10.5 \text{ kJ} + [(-1) \times 8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{mol}^{-1} \times 298 \text{ K}]$\n$\Delta H^o = -10.5 \text{ kJ} - 2.48 \text{ kJ} = -12.98 \text{ kJ}$\n$\Delta H^o = -12.98 \times 10^3 \text{ J}$\n\n$\text{Step 2: Calculate } \Delta G^o \text{ for given reaction}$\n\n$\text{We know that,}$\n\n$\Delta G^o = \Delta H^o - T \Delta S^o$\n$\Delta G^o = -12.98 \times 10^3 \text{ J} - 298 \text{ K}(-44 \text{ J K}^{-1})$\n$\Delta G^o = -12980 \text{ J} + 13112 \text{ J}$\n$\Delta G^o = 132 \text{ J}$\n$\Delta G^o = +0.132 \text{ kJ} \approx +0.16 \text{ kJ}$

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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
}