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Current Question (ID: 18354)

Question:

$\text{Consider the first-order gas-phase decomposition reaction given below.}$ $\text{A(g)} \rightarrow \text{B(g)} + \text{C(g)}$ $\text{The initial pressure of the system before the decomposition of A was } P_i. \text{ After the lapse of time } t, \text{ the total pressure of the system increased by } X \text{ units and became } P_t. \text{ The rate constant } k \text{ for the reaction is:}$

Options:

1. $k = \frac{2.303}{t} \log \frac{P_i}{P_i - x}$
2. $k = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}$
3. $k = \frac{2.303}{t} \log \frac{P_i}{2P_i + P_t}$
4. $k = \frac{2.303}{t} \log \frac{P_i}{P_i + x}$

Solution:

$\text{HINT: Use First order gas-phase reaction}$ $\text{A(g)} \rightarrow \text{B(g)} + \text{C(g)}$ $P_i \quad 0 \quad 0$ $P_i - x \quad x \quad x$ $\text{For the first order reaction}$ $P_t = P_i - x + x = P_i + x$ $x = P_t - P_i$ $k = \frac{2.303}{t} \log \frac{P_i}{P_i - x}$ $= \frac{2.303}{t} \log \frac{P_i}{P_i - (P_t - P_i)}$ $= \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}$ $\text{According to the problem, the initial pressure of the system is } P_i, \text{ and after a certain time } t, \text{ the total pressure of the system increased by } X \text{ units and became } P_t. \text{ As } X \text{ is partial pressure and we write our final expression in initial pressure and final pressure so most suitable answer will be the 2nd one not 1st as it uses the term } X \text{ as the it is a partial pressure.}$ $\text{Hence, option second is the correct answer.}$

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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
}