Import Question JSON

Current Question (ID: 18373)

Question:

$\text{The kinetic data for the reaction: } 2\text{A} + \text{B}_2 \rightarrow 2\text{AB} \text{ are as given below}$ $\begin{array}{ccc} [\text{A}]/\text{mol L}^{-1} & [\text{B}_2]/\text{mol L}^{-1} & \text{Rate}/\text{mol L}^{-1}\text{s}^{-1} \\ 0.5 & 1.0 & 2.5 \times 10^{-3} \\ 1.0 & 1.0 & 5.0 \times 10^{-3} \\ 0.5 & 2.0 & 1 \times 10^{-2} \end{array}$ $\text{The order of reaction with respect to A and B}_2 \text{ is, respectively:}$

Options:

1. $1 \text{ and } 2$
2. $2 \text{ and } 1$
3. $1 \text{ and } 1$
4. $2 \text{ and } 2$

Solution:

$\text{HINT: Use general expression of rate law and find the order w.r.t. individual reactant.}$ $\text{Explanation:}$ $2.5 \times 10^{-3} = K[0.5]^{\alpha}[1.0]^{\beta} \quad \ldots(1)$ $5 \times 10^{-3} = K[1.0]^{\alpha}[1.0]^{\beta} \quad \ldots(2)$ $1 \times 10^{-2} = K[0.5]^{\alpha}[2.0]^{\beta} \quad \ldots(3)$ $\text{Dividing equation (1) and (2)}$ $\frac{1}{2} = \left[ \frac{1}{2} \right]^{\alpha} ; \text{ hence } \alpha = 1$ $\text{Dividing equation (1) and (3)}$ $\frac{2.5 \times 10^{-3}}{1 \times 10^{-2}} = \left( \frac{1.0}{2.0} \right)^{\beta} = \frac{1}{4} = \left( \frac{1}{2} \right)^{\beta} ; \beta = 2$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}