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Current Question (ID: 21520)

Question:
$\text{Consider the given elementary reaction:}$ $\text{A (g) + B (g)} \rightarrow \text{C (g) + D (g)}$ $\text{If the volume of the reaction mixture is suddenly reduced to } \frac{1}{3} \text{ of its initial volume, the reaction rate will become 'x' times the original reaction rate. The value of x is:}$
Options:
  • 1. $\frac{1}{9}$
  • 2. $9$
  • 3. $\frac{1}{3}$
  • 4. $3$
Solution:
$\text{Hint: } R = K[\text{A}]^1[\text{B}]^1$ $\text{Explanation:}$ $R_1 = K[\text{A}]^1[\text{B}]^1$ $R_1 = K\left[\frac{n_A}{V}\right]^1\left[\frac{n_B}{V}\right]^1$ $R_2 = K\left[\frac{3n_A}{V}\right]^1\left[\frac{3n_B}{V}\right]^1$ $R_2 = 9R_1$

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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
}