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Current Question (ID: 18500)

Question:

$\text{The correct statement among the following is:}$

Options:

1. $\text{In the decomposition of oxide into oxygen and gaseous metal, entropy increases.}$
2. $\text{Decomposition of oxide is an endothermic change.}$
3. $\text{To make } \Delta G^\circ \text{ negative, the temperature should be high enough so that } T \Delta S^\circ > \Delta H^\circ.\text{}$
4. $\text{All of the above.}$

Solution:

$\text{1. When an oxide of any metal is heated the oxide decomposes to form pure metal and oxygen gas.}$ $\text{The reaction is as follows: } \text{MO (s)} \rightarrow \text{M(s)} + \text{O}_2 \text{(g)}$ $\text{Entropy is the measure of randomness. The entropy of solids is very low because in solids the atoms or ions or molecules have an ordered arrangement.}$ $\text{The entropy of gases is higher because the constituents of gases are not in an ordered arrangement. In the reaction, a solid reactant is converted to two gaseous products.}$ $\text{So, the entropy increases as the reaction proceeds.}$ $\text{Thus, the statement in the decomposition of an oxide, into oxygen and gaseous metal, entropy increases is correct.}$ $\text{2. When an oxide of any metal is heated, the oxide decomposes to form pure metal and oxygen gas.}$ $\text{The reaction is as follows: } \text{MO (s)} \rightarrow \text{M(s)} + \text{O}_2 \text{(g)}$ $\text{In the reaction, we can see that the bond between the metal and oxygen breaks.}$ $\text{A large amount of energy is needed to break this bond. This energy is absorbed from the surrounding. We know that the reactions in which energy is absorbed are known as endothermic reactions.}$ $\text{Thus decomposition of oxide is an endothermic reaction. Thus the statement 'decomposition of oxide is an endothermic change' is also correct.}$ $\text{3. We know that the expression that gives the relation between free energy, entropy and enthalpy is as follows:}$ $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$ $\text{Where } \Delta G^\circ \text{ is the change in Gibb's free energy. T is the temperature. } \Delta S^\circ \text{ is the change in entropy.}$ $\text{When temperature is high } T \Delta S^\circ > \Delta H^\circ. \text{ And thus, } \Delta G^\circ \text{ becomes negative. Thus the statement 'to make } \Delta G^\circ \text{ negative temperature should be high enough so that } T \Delta S^\circ > \Delta H^\circ \text{ is correct.}$ $\text{(1), (2) and (3) statements are correct.}$

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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
}