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Current Question (ID: 18326)

Question:
$\text{For a first-order reaction } A \rightarrow P, \text{ rate constant } (k) \text{ [dependent on temperature (T)] was found to follow the equation } \log k = \left(-2000\right) \frac{1}{T} + 6.0. \text{ The pre-exponential factor } A \text{ and the activation energy } E_a, \text{ respectively, are:}$
Options:
  • 1. $1.0 \times 10^6 \text{ s}^{-1} \text{ and } 9.2 \text{ kJ mol}^{-1}$
  • 2. $6.0 \text{ s}^{-1} \text{ and } 16.6 \text{ kJ mol}^{-1}$
  • 3. $1.0 \times 10^6 \text{ s}^{-1} \text{ and } 16.6 \text{ kJ mol}^{-1}$
  • 4. $1.0 \times 10^6 \text{ s}^{-1} \text{ and } 38.3 \text{ kJ mol}^{-1}$
Solution:
$\text{HINT: Use Arrhenius Equation.}$ $\text{Explanation:}$ $\ln k = -\frac{2000}{T} + 6.0$ $\log_{10} k = -\frac{E_a}{2.303 \cdot RT} + \log_{10} A$ $\therefore \frac{E_a}{2.303} \cdot R = 2000$ $\text{or } E_a = 2.303 \times 2000 \times 8.314$ $= 38.3 \times 10^3 \text{ J} = 38.3 \text{ kJ}$ $\text{Also, } A = 10^6$

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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
}