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Current Question (ID: 18287)

Question:

$\text{The decomposition of a gaseous compound yields the following information:}$ $\begin{array}{|c|c|c|} \hline \text{Initial pressure, atm} & 1.6 & 0.8 & 0.4 \\ \hline \text{Time for 50 \% reaction, min} & 80 & 113 & 160 \\ \hline \end{array}$ $\text{The order of the reaction will be:}$

Options:

1. $1.0$
2. $1.5$
3. $2.0$
4. $0.5$

Solution:

$\text{HINT: } t_{1/2} \propto \frac{1}{p^{n-1}}$ $\text{Explanation:}$ $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{80}{113} = \left(\frac{p_2}{p_1}\right)^{n-1}$ $\text{where } p_2, \ p_1 \text{ are the initial pressures}$ $\frac{80}{113} = \left(\frac{0.8}{1.6}\right)^{n-1}$ $\text{Taking logarithms.}$ $\log 0.7 = (n-1) \log 0.5$ $\text{solving, } n=1.5$

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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
}