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Current Question (ID: 8261)

Question:

$\text{For the following given equations and } \Delta\text{H}^{\circ} \text{ values, determine the enthalpy of reaction at 298 K for the reaction:}$ $\text{C}_2\text{H}_4(g) + 6\text{F}_2(g) \rightarrow 2\text{CF}_4(g) + 4\text{HF}(g)$ $\text{H}_2(g) + \text{F}_2(g) \rightarrow 2\text{HF}(g) \quad \Delta\text{H}^{\circ}_1 = -537 \text{ kJ}$ $\text{C}(s) + 2\text{F}_2(g) \rightarrow \text{CF}_4(g) \quad \Delta\text{H}^{\circ}_2 = -680 \text{ kJ}$ $2\text{C}(s) + 2\text{H}_2(g) \rightarrow \text{C}_2\text{H}_4(g) \quad \Delta\text{H}^{\circ}_3 = 52 \text{ kJ}$

Options:

1. $-1165 \text{ kJ}$
2. $-2486 \text{ kJ}$
3. $+1165 \text{ kJ}$
4. $+2486 \text{ kJ}$

Solution:

$\text{Hint: Hess's Law of Constant Heat Summation}$ $\text{H}_2(g) + \text{F}_2(g) \rightarrow 2\text{HF}(g) \quad \Delta\text{H}^{\circ}_1 = -537 \text{ kJ} \quad ......(i)$ $\text{C}(s) + 2\text{F}_2(g) \rightarrow \text{CF}_4(g) \quad \Delta\text{H}^{\circ}_2 = -680 \text{ kJ} \quad ......(ii)$ $2\text{C}(s) + 2\text{H}_2(g) \rightarrow \text{C}_2\text{H}_4(g) \quad \Delta\text{H}^{\circ}_3 = 52 \text{ kJ} \quad ......(iii)$ $\text{Consider the above reaction.}$ $\text{Multiple equations 1 and 2 with 2. Add both the equation. Subtract equation third from the resulting equation.}$ $\Delta\text{H}^{\circ} = 2 \Delta \text{H}^{\circ}_1 + 2 \times \Delta \text{H}^{\circ}_2 - \Delta\text{H}^{\circ}_3$ $\Delta\text{H}^{\circ} = ((2(-537)) + (2(-680)) - 52) \text{ kJ}$ $\Delta\text{H}^{\circ} = (- 1074 - 1360 - 52) \text{ kJ}$ $\Delta\text{H}^{\circ} = -2486 \text{ kJ}$

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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
}