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Current Question (ID: 8321)

Question:
$\text{What is the value of } \Delta\text{H for the reaction } 2\text{Al} + \text{Cr}_2\text{O}_3 \rightarrow 2\text{Cr} + \text{Al}_2\text{O}_3\text{, knowing that the enthalpies of}$ $\text{formation of Al}_2\text{O}_3 \text{ and Cr}_2\text{O}_3 \text{ are } -1596 \text{ kJ and } -1134 \text{ kJ, respectively?}$
Options:
  • 1. $-1365 \text{ kJ}$
  • 2. $2730 \text{ kJ}$
  • 3. $-2730 \text{ kJ}$
  • 4. $-462 \text{ kJ}$
Solution:
$\text{Hint: } \Delta\text{H}^0_r = [2 \Delta \text{H}_f] (\text{Cr}) + \Delta\text{H}_f (\text{Al}_2 \text{O}_3) - [2 \Delta \text{H}_f] (\text{Al}) + \Delta\text{H}_f (\text{Cr}_2 \text{O}_3)$ $\text{Step 1: Identify the given information}$ $\text{The reaction is: } 2\text{Al} + \text{Cr}_2\text{O}_3 \rightarrow 2\text{Cr} + \text{Al}_2\text{O}_3$ $\text{Given:}$ $\Delta\text{H}_f(\text{Al}_2\text{O}_3) = -1596 \text{ kJ}$ $\Delta\text{H}_f(\text{Cr}_2\text{O}_3) = -1134 \text{ kJ}$ $\text{Step 2: Apply the formula for calculating the enthalpy change of a reaction}$ $\text{The standard enthalpy change for a reaction can be calculated using the following formula:}$ $\Delta\text{H}^0_r = \sum \Delta\text{H}_f(\text{products}) - \sum \Delta\text{H}_f(\text{reactants})$ $\text{For our reaction:}$ $\Delta\text{H}^0_r = [2 \times \Delta\text{H}_f(\text{Cr}) + \Delta\text{H}_f(\text{Al}_2\text{O}_3)] - [2 \times \Delta\text{H}_f(\text{Al}) + \Delta\text{H}_f(\text{Cr}_2\text{O}_3)]$ $\text{Step 3: Substitute the known values and calculate}$ $\text{Remember that the enthalpy of formation of elements in their standard states is zero by definition:}$ $\Delta\text{H}_f(\text{Cr}) = 0 \text{ kJ}$ $\Delta\text{H}_f(\text{Al}) = 0 \text{ kJ}$ $\text{Substituting these values:}$ $\Delta\text{H}^0_r = [2 \times 0 + (-1596 \text{ kJ})] - [2 \times 0 + (-1134 \text{ kJ})]$ $\Delta\text{H}^0_r = -1596 \text{ kJ} - (-1134 \text{ kJ})$ $\Delta\text{H}^0_r = -1596 \text{ kJ} + 1134 \text{ kJ}$ $\Delta\text{H}^0_r = -462 \text{ kJ}$ $\text{Therefore, the enthalpy change for the reaction is } -462 \text{ kJ}$ $\text{This negative value indicates that the reaction is exothermic, releasing 462 kJ of energy.}$

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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
}