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Current Question (ID: 18388)

Question:

$\text{An incorrect statement regarding the catalyst is:}$

Options:

1. $\text{It catalyzes the forward and backward reactions to the same extent.}$
2. $\text{It alters Gibbs free energy of the reaction.}$
3. $\text{It is a substance that does not change the equilibrium constant of a reaction.}$
4. $\text{It provides an alternate mechanism by reducing activation energy between reactants and products.}$

Solution:

$\text{HINT: Catalyst - Any substance that alter the rate of a reaction without itself being consumed.}$ $\text{Explanation:}$ $\text{Characteristics of catalyst is as follows:}$ $\text{(1) It catalyzes the forward and backward reaction to the same extent as it decreases the energy of activation hence, increases the rate of both the reactions.}$ $\text{(2) Since, reaction quotient is the relation between the concentration of reactants and products. Hence, the catalyst does not alter Gibbs free energy as it is related to the reaction quotient. Thus, Gibbs's free energy does not change during the reaction when the catalyst is added to it.}$ $\Delta G = -RT \ln Q$ $\text{where, Q = reaction quotient}$ $\text{(3) It doesn't alter the equilibrium of reaction as the equilibrium constant is also a concentration-dependent term.}$ $\text{(4) It provides an alternate mechanism by reducing activation energy between reactants and products.}$

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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
}