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Current Question (ID: 21487)

Question:

$\text{For the reaction, aA + bB} \rightarrow \text{cC + dD the plot of } \log k \text{ vs } \frac{1}{T} \text{ is given below:}$ $\text{Find the temperature(K) at which the rate constant of the reaction is } 10^{-4}\text{s}^{-1}?$ $\text{(Rounded-off to the nearest integer)}$ $\text{[Given: The rate constant of the reaction is } 10^{-5} \text{ s}^{-1} \text{ at 500 K.]}$

Options:

1. $546 \text{ K}$
2. $536 \text{ K}$
3. $516 \text{ K}$
4. $526 \text{ K}$

Solution:

$\text{Hint: Use Arrhenius equation, that is, } \log K = \log A - \frac{E_a}{2.303RT}$ $\text{Step 1: The Arrhenius equation is as follows:}$ $\log K = \log A - \frac{E_a}{2.303RT}$ $\text{From the equation,}$ $\text{Slope} = \frac{E_a}{2.303R} = 10,000$ $E_a = 10000 \times 2.303R$ $\text{Step 2:}$ $\text{Calculate the temperature at which the rate constant of the reaction is } 10^{-4}\text{s}^{-1} \text{ as follows:}$ $\log \left( \frac{k_2}{k_1} \right) = \frac{10000 \times 2.303R}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$ $\log \left( \frac{10^{-4}}{10^{-5}} \right) = 10,000 \left[ \frac{1}{500} - \frac{1}{T_2} \right]$ $\frac{1}{10000} = \frac{1}{500} - \frac{1}{T_2}$ $\frac{1}{10000} - \frac{1}{500} = -\frac{1}{T_2}$ $\frac{500 - 10000}{5 \times 10^6} = -\frac{1}{T_2}$ $0.0019 = \frac{1}{T_2}$ $T_2 = 526 \text{ K}$ $\text{Hence, option 4 is the correct answer.}$

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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
}