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Current Question (ID: 8435)

Question:

$\text{Given the reaction:}$ $\text{CH}_3\text{COOH}_{(l)} + \text{C}_2\text{H}_5\text{OH}_{(l)} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5_{(l)} + \text{H}_2\text{O}_{(l)}$ $\text{At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. The equilibrium constant of the reaction will be:}$

Options:

1. $0.78$
2. $1.3$
3. $2.47$
4. $3.92$

Solution:

$\text{HINT: } K_c = \frac{[\text{CH}_3 \text{COOC}_2 \text{H}_5] [\text{H}_2\text{O}]}{[\text{CH}_3 \text{COOH}] [\text{C}_2\text{H}_5 \text{OH}]}$ $\text{Explanation:}$ $\text{Step 1: Write down the equilibrium concentration of reactants and products}$ $\text{Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.}$ $\text{The given reaction is:}$ $\text{CH}_3 \text{COOH}_{(l)} + \text{C}_2\text{H}_5 \text{OH}_{(l)} \rightleftharpoons \text{CH}_3 \text{COOC}_2 \text{H}_5_{(l)} + \text{H}_2\text{O}_{(l)}$ $\text{Initial conc.} \quad \frac{1}{V}\text{M} \quad \frac{0.18}{V}\text{M} \quad 0 \quad 0$ $\text{At equilibrium} \quad \frac{1-0.171}{V} \quad \frac{0.18-0.171}{V} \quad \frac{0.171}{V}\text{M} \quad \frac{0.171}{V}\text{M}$ $\text{Step 2: Calculate equilibrium constant for the given reaction}$ $\text{Therefore, the equilibrium constant for the given reaction is:}$ $K_c = \frac{[\text{CH}_3 \text{COOC}_2 \text{H}_5] [\text{H}_2\text{O}]}{[\text{CH}_3 \text{COOH}] [\text{C}_2\text{H}_5 \text{OH}]}$ $= \frac{\frac{0.171}{V} \times \frac{0.171}{V}}{\frac{0.829}{V} \times \frac{0.009}{V}} = 3.919$ $= 3.92 \text{ (approx.)}$

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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
}