Import Question JSON

« Back to Question List Edit Question »

Current Question (ID: 8319)

Question:
$\text{The standard heat of combustion of propane is } -2220.1 \text{ kJ mol}^{-1}\text{. The standard heat of vaporisation of liquid water is } 44.0 \text{ kJ mol}^{-1}\text{. The enthalpy change for the reaction is−}$ $\text{C}_3\text{H}_8 \text{ (g)} + 5\text{O}_2 \text{ (g)} \rightarrow 3\text{CO}_2 \text{ (g)} + 4\text{H}_2\text{O(g)}$
Options:
  • 1. $-2220.1 \text{ kJ}$
  • 2. $-2044.1 \text{ kJ}$
  • 3. $-2396.1 \text{ kJ}$
  • 4. $-2176.1 \text{ kJ}$
Solution:
$\text{Hint: } \Delta\text{H}^{\circ}_{(\text{Rxn})} = \Delta\text{H}^{\circ}_{\text{comb}} (\text{C}_3\text{H}_8) + \Delta\text{H}^{\circ}_{\text{vap}} (4\text{H}_2\text{O})$ $\text{Step 1: Analyze the information given and identify what we need to calculate}$ $\text{We are given:}$ $\text{- Standard heat of combustion of propane: } \Delta\text{H}^{\circ}_{\text{comb}} (\text{C}_3\text{H}_8) = -2220.1 \text{ kJ mol}^{-1}$ $\text{- Standard heat of vaporization of water: } \Delta\text{H}^{\circ}_{\text{vap}} (\text{H}_2\text{O}) = 44.0 \text{ kJ mol}^{-1}$ $\text{We need to find the enthalpy change for the reaction:}$ $\text{C}_3\text{H}_8 \text{ (g)} + 5\text{O}_2 \text{ (g)} \rightarrow 3\text{CO}_2 \text{ (g)} + 4\text{H}_2\text{O(g)}$ $\text{Step 2: Understand the relationship between the given values and the required enthalpy change}$ $\text{The standard heat of combustion of propane typically refers to the reaction where water is produced in the liquid state:}$ $\text{C}_3\text{H}_8 \text{ (g)} + 5\text{O}_2 \text{ (g)} \rightarrow 3\text{CO}_2 \text{ (g)} + 4\text{H}_2\text{O(l)}$ $\text{However, in our target reaction, water is produced in the gaseous state.}$ $\text{To get from liquid water to gaseous water, we need to account for the heat of vaporization.}$ $\text{Step 3: Calculate the enthalpy change for the reaction}$ $\text{First, let's calculate the total heat of vaporization for 4 moles of water:}$ $\Delta\text{H}^{\circ}_{\text{vap}} \text{ of 4 mol H}_2\text{O} = 4 \times 44.0 \text{ kJ mol}^{-1} = 176.0 \text{ kJ}$ $\text{Now, we can calculate the enthalpy change for the target reaction:}$ $\Delta\text{H}^{\circ}_{(\text{Rxn})} = \Delta\text{H}^{\circ}_{\text{comb}} (\text{C}_3\text{H}_8) + \Delta\text{H}^{\circ}_{\text{vap}} (4\text{H}_2\text{O})$ $= -2220.1 \text{ kJ} + 176.0 \text{ kJ}$ $= -2044.1 \text{ kJ}$ $\text{Therefore, the enthalpy change for the reaction is } -2044.1 \text{ kJ}$ $\text{This makes sense because less energy is released when water is produced as a gas compared to when it's produced as a liquid, since energy is required to vaporize water.}$

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
}