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Current Question (ID: 8274)

Question:

$\text{Determine the enthalpy change for the specified reaction:}$ $2\text{H}_2\text{O}_2(l) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g).$ $(\text{Given the heat of formation of H}_2\text{O}_2(l) \text{ and H}_2\text{O}(l) \text{ are } -188 \text{ and } -286 \text{ kJ/mol respectively})$

Options:

1. $-196 \text{ kJ/mol}$
2. $+196 \text{ kJ/mol}$
3. $+948 \text{ kJ/mol}$
4. $-948 \text{ kJ/mol}$

Solution:

$\text{Hint: Total enthalpy change for product - total enthalpy change for reactant}$ $\text{For the reaction:}$ $2\text{H}_2\text{O}_2(l) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g) \Delta\text{H}=?$ $\text{We can calculate the enthalpy change using the standard heats of formation:}$ $\Delta\text{H} = [(2 \times \Delta\text{H}_f \text{ of H}_2\text{O}(l)) + (\Delta\text{H}_f \text{ of O}_2(g))]$ $\quad \quad - [2 \times \Delta\text{H}_f \text{ of H}_2\text{O}_2(l)]$ $\text{Note that the heat of formation of elemental oxygen } \text{O}_2(g) \text{ in its standard state is 0 kJ/mol.}$ $\Delta\text{H} = [(2 \times -286) + (0)] - [2 \times -188]$ $\Delta\text{H} = [-572 + 376]$ $\Delta\text{H} = -196 \text{ kJ/mol}$ $\text{The negative value of enthalpy change indicates that this is an exothermic reaction, where heat is released to the surroundings.}$

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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
}