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Question:
$\text{Determine the enthalpy of formation for } \text{H}_2\text{O}_2\text{(l), using the following enthalpies of reaction:}$ $\text{N}_2\text{H}_4\text{(l)}+2\text{H}_2\text{O}_2\text{(l)}\rightarrow \text{N}_2\text{(g)}+4\text{H}_2\text{O}\text{(l)}; \Delta \text{H}_1^0 = -818 \text{ kJ/mol}$ $\text{N}_2\text{H}_4\text{(l)}+\text{O}_2\text{(g)}\rightarrow \text{N}_2\text{(g)}+2\text{H}_2\text{O}\text{(l)}; \Delta \text{H}_2^0 = -622 \text{ kJ/mol}$ $\text{H}_2\text{(g)}+\frac{1}{2}\text{O}_2\text{(g)}\rightarrow \text{H}_2\text{O}\text{(l)}; \Delta \text{H}_3^0 = -285 \text{ kJ/mol}$
Options:
  • 1. $-383 \text{ kJ/mol}$
  • 2. $-187 \text{ kJ/mol}$
  • 3. $-49 \text{ kJ/mol}$
  • 4. $\text{None of the above}$
Solution:
$\text{Hint: } \Delta_r\text{H}^{\circ} = \sum \text{a}_i\Delta\text{H}_f^0(\text{products}) - \sum \text{b}_i\Delta\text{H}_f^0(\text{reactants})$ $\text{Step 1: Identify the formation reaction for H}_2\text{O}_2\text{(l)}$ $\text{The formation reaction for hydrogen peroxide is:}$ $\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{H}_2\text{O}_2\text{(l)}$ $\text{Step 2: Use the given information to determine the enthalpy of formation}$ $\text{We need to combine the given reactions in a way that gives us the formation reaction for H}_2\text{O}_2\text{(l).}$ $\text{Let's try:}$ $\text{1. Reverse equation 1 (multiplying by -1):}$ $\text{N}_2\text{(g)}+4\text{H}_2\text{O}\text{(l)} \rightarrow \text{N}_2\text{H}_4\text{(l)}+2\text{H}_2\text{O}_2\text{(l)}; \Delta\text{H} = +818 \text{ kJ/mol}$ $\text{2. Use equation 2 as is:}$ $\text{N}_2\text{H}_4\text{(l)}+\text{O}_2\text{(g)}\rightarrow \text{N}_2\text{(g)}+2\text{H}_2\text{O}\text{(l)}; \Delta\text{H} = -622 \text{ kJ/mol}$ $\text{3. Multiply equation 3 by 2:}$ $2\text{H}_2\text{(g)}+\text{O}_2\text{(g)}\rightarrow 2\text{H}_2\text{O}\text{(l)}; \Delta\text{H} = 2 \times (-285) = -570 \text{ kJ/mol}$ $\text{Now, combine these equations:}$ $\text{Reversed equation 1 + equation 2 + 2 × equation 3:}$ $\text{N}_2\text{(g)}+4\text{H}_2\text{O}\text{(l)} + \text{N}_2\text{H}_4\text{(l)}+\text{O}_2\text{(g)} + 2\text{H}_2\text{(g)}+\text{O}_2\text{(g)} \rightarrow \text{N}_2\text{H}_4\text{(l)}+2\text{H}_2\text{O}_2\text{(l)} + \text{N}_2\text{(g)}+2\text{H}_2\text{O}\text{(l)} + 2\text{H}_2\text{O}\text{(l)}$ $\text{Simplifying:}$ $4\text{H}_2\text{O}\text{(l)} + \text{O}_2\text{(g)} + 2\text{H}_2\text{(g)}+\text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O}_2\text{(l)} + 4\text{H}_2\text{O}\text{(l)}$ $\text{Further simplifying:}$ $2\text{H}_2\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O}_2\text{(l)}$ $\text{This gives us:}$ $\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{H}_2\text{O}_2\text{(l)}$ (for 1 mole of H}_2\text{O}_2\text{)} $\text{Step 3: Calculate the enthalpy change}$ $\Delta\text{H}_f^0 = (-\Delta\text{H}_1^0) + (\Delta\text{H}_2^0) + (2\Delta\text{H}_3^0)$ $= \{-(-818)\} + (-622) + (2 \times (-285))$ $= 818 - 622 - 570$ $= -374 \text{ kJ/mol}$ $\text{This value is for 2 moles of H}_2\text{O}_2\text{(l), so for 1 mole:}$ $\Delta\text{H}_f = \frac{-374}{2} = -187 \text{ kJ/mol}$ $\text{Therefore, the enthalpy of formation for H}_2\text{O}_2\text{(l) is } -187 \text{ kJ/mol}$

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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
}