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Current Question (ID: 21039)

Question:

$\text{The enthalpy changes for the following processes are listed below:}$ $\text{Cl}_2(g) \rightarrow 2\text{Cl}(g), 242.3 \text{ kJ mol}^{-1}$ $\text{I}_2(g) \rightarrow 2\text{I}(g), 151.0 \text{ kJ mol}^{-1}$ $\text{ICl}(g) \rightarrow \text{I}(g) + \text{Cl}(g), 211.3 \text{ kJ mol}^{-1}$ $\text{I}_2(s) \rightarrow \text{I}_2(g), 62.76 \text{ kJ mol}^{-1}$ $\text{Given that the standard states for iodine and chlorine are } \text{I}_2(s) \text{ and } \text{Cl}_2(g), \text{ the standard enthalpy of formation of ICl (g) is:}$

Options:

1. $-14.6 \text{ kJ mol}^{-1}$
2. $-20.8 \text{ kJ mol}^{-1}$
3. $+16.8 \text{ kJ mol}^{-1}$
4. $+244.8 \text{ kJ mol}^{-1}$

Solution:

$\text{Hint: Use Hess's law}$ $\text{Step 1:}$ $\text{The ICl formation reaction is as follows:}$ $\frac{1}{2} \text{I}_2(s) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{ICl}(g)$ $\text{first, multiple equations 1 and 2 by } \frac{1}{2}$ $\frac{1}{2} \text{Cl}_2(g) \rightarrow \text{Cl}(g), 121.15 \text{ kJ mol}^{-1} \quad .....(5)$ $\frac{1}{2} \text{I}_2(g) \rightarrow \text{I}(g), 75.5 \text{ kJ mol}^{-1} \quad .....(6)$ $\text{Reverse equation third}$ $\text{I}(g) + \text{Cl}(g) \rightarrow \text{ICl}(g) -211.3 \text{ kJ mol}^{-1} \quad .....(7)$ $\text{Multiple equation four by } \frac{1}{2}$ $\frac{1}{2} \text{I}_2(s) \rightarrow \frac{1}{2} \text{I}_2(g), 31.38 \text{ kJ mol}^{-1} \quad .....(8)$ $\text{Step 2:}$ $\text{Add equations 5, 6, 7, and 8, and also add their enthalpy so that the ICl formation reaction is obtained.}$ $\Delta H_f, \text{ICl} = 121.15 + 75.5 + (-211.3) + 31.38$ $= 16.8 \text{ kJ mol}^{-1}$

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{
  "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
}