Import Question JSON

Current Question (ID: 8302)

Question:

$\text{N}_2 + 3 \text{ H}_2 \rightarrow 2\text{NH}_3; \quad \Delta_\text{r} \text{H}^{\circ} = -92.4 \text{ kJ mol}^{-1}. \text{ The standard enthalpy of formation of NH}_3 \text{ gas in the above reaction would be:}$

Options:

1. $-92.4 \text{ J (mol)}^{-1}$
2. $-46.2 \text{ kJ (mol)}^{-1}$
3. $+46.2 \text{ J (mol)}^{-1}$
4. $+92.4 \text{ kJ (mol)}^{-1}$

Solution:

$\text{Hint: Standard enthalpy of formation of NH}_3 = \frac{1}{2} \Delta_\text{r}\text{H}$ $\text{The standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.}$ $\text{Step 1: Write down the given information}$ $\text{N}_2 + 3 \text{ H}_2 \rightarrow 2\text{NH}_3; \quad \Delta_\text{r} \text{H}^{\circ} = -92.4 \text{ kJ mol}^{-1}$ $\text{Step 2: Write down the reaction for the formation of 1 mole of NH}_3$ $\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{NH}_3(g)$ $\text{Step 3: Calculate the standard enthalpy of formation of NH}_3 \text{ gas}$ $\text{Standard enthalpy of formation of NH}_3 = \frac{1}{2} \Delta_\text{r} \text{H}^{\circ}$ $= \frac{1}{2}(-92.4 \text{ kJ mol}^{-1}) = -46.2 \text{ kJ mol}^{-1}$

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
}