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Current Question (ID: 18265)

Question:
$\text{A reaction is first-order with respect to A and second-order with respect to B.}$ $\text{The concentration of B is increased three times. The new rate of the reaction would:}$
Options:
  • 1. $\text{Decrease 9 times}$
  • 2. $\text{Increase 9 times}$
  • 3. $\text{Increase 6 times}$
  • 4. $\text{Decrease 6 times}$
Solution:
$\text{Hint: Use rate law.}$ $\text{Step 1:}$ $\text{The differential rate equation will be}$ $-\frac{d[\text{R}]}{dt} = \text{K[A][B]}^2$ $\text{Step 2:}$ $\text{If the concentration of B is increased three times, then}$ $-\frac{d[\text{R}]}{dt} = \text{K[A][3B]}^2$ $= 9 \cdot \text{K[A][B]}^2$ $\text{Therefore, the rate of reaction will increase 9 times.}$

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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
}