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Current Question (ID: 8268)

Question:

$\text{For the given reaction}$ $2\text{H}_2\text{O}_2(l) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g), \text{ the heat of formations of H}_2\text{O}_2(l) \text{ and H}_2\text{O (l) are -188 kJ/mol \& -286 kJ/mol respectively. The change in the enthalpy of the reaction will be:}$

Options:

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

Solution:

$\text{Hint: } \Delta\text{H}_r = (2 \, \Delta \text{H}_f) , \text{H}_2\text{O} + \Delta\text{H}_f, \text{O}_2 - \Delta\text{H}_f, \text{H}_2\text{O}_2$ $\text{The following general equation can be used for the enthalpy change calculation.}$ $\Delta_r H^{\Theta} = \sum_i a_i \Delta_f H^{\Theta} \text{(products)} - \sum_i b_i \Delta_f H^{\Theta} \text{(reactants)}$ $\text{Calculate the value of change in the enthalpy of the reaction as follows:}$ $\Delta\text{H}_r = [2 \, \Delta \text{H}_f(\text{H}_2\text{O})] + \Delta\text{H}_f(\text{O}_2) - [2 \times \Delta\text{H}_f(\text{H}_2\text{O}_2)]$ $\Delta\text{H}_r = [2 \times (-286)] + 0 - [2 \times (-188)]$ $\Delta\text{H}_r = -572 + 376$ $\Delta\text{H}_r = -196 \text{ kJ mol}^{-1}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}