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Current Question (ID: 18329)

Question:

$\text{For a chemical reaction } A \rightarrow \text{product, the postulated mechanism of the reaction is as follows.}$ $A \overset{k_1}{\underset{k_2}{\rightleftharpoons}} 3B \overset{k_3}{\rightarrow} C \text{ (R.D.S)}$ $\text{If the rate constants for individual reactions are } k_1, k_2 \text{ and } k_3, \text{ and activation energies are}$ $E_{a1} = 180 \text{ kJ mol}^{-1},$ $E_{a2} = 90 \text{ kJ mol}^{-1},$ $E_{a3} = 40 \text{ kJ mol}^{-1}$ $\text{then overall activation energy for the reaction given above is}$

Options:

1. $70 \text{ kJ mol}^{-1}$
2. $-10 \text{ kJ mol}^{-1}$
3. $310 \text{ kJ mol}^{-1}$
4. $130 \text{ kJ mol}^{-1}$

Solution:

$\text{HINT: Use Arrhenius Equation.}$ $\text{Explanation:}$ $A \overset{k_1}{\underset{k_2}{\rightleftharpoons}} 3B \overset{k_3}{\rightarrow} C$ $\text{Rate } = k_3[B]^3 \quad \text{.......(1)}$ $\text{Find the value of } [B]^3 \text{ and substitute in equation 1.}$ $K = \frac{[B]^3}{[A]} ; [B]^3 = \left( \frac{k_1}{k_2} \right) [A]$ $\text{Rate } = k_3 \times \frac{k_1}{k_2} [A]$ $\text{The overall rate constant is as follows:}$ $k = k_3 \times \frac{k_1}{k_2}$ $\text{Where}$ $A e^{-E/RT} = \left( \frac{A_1 e^{-E_1/RT}}{A_2 e^{-E_2/RT}} \right) \left( A_3 e^{-E_3/RT} \right)$ $E = E_1 - E_2 + E_3 = 130.0 \text{ kJ mol}^{-1}$

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{
  "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
}