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Current Question (ID: 8352)

Question:
$\text{An enthalpy diagram for a particular reaction is given below:}$ $\text{[Enthalpy diagram showing reactants at energy level H}_r\text{, products at energy level H}_p\text{, with }\Delta_f\text{H indicated as net heat absorbed from surroundings. The diagram shows H}_p > \text{H}_r\text{]}$ $\text{The correct statement among the following is-}$ $\text{1. Reaction is spontaneous}$ $\text{2. Reaction is non-spontaneous}$ $\text{3. Cannot predict spontaneity of the reaction from the graph given above}$ $\text{4. None of the above}$
Options:
  • 1. $\text{Reaction is spontaneous}$
  • 2. $\text{Reaction is non-spontaneous}$
  • 3. $\text{Cannot predict spontaneity of the reaction from the graph given above}$
  • 4. $\text{None of the above}$
Solution:
$\text{Hint:}$ $\Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S}$ $\text{To determine whether a reaction is spontaneous or not, we need to analyze the Gibbs free energy change (}\Delta\text{G) for the reaction:}$ $\Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S}$ $\text{For a reaction to be spontaneous, }\Delta\text{G must be negative.}$ $\text{From the enthalpy diagram provided:}$ $\text{1. We can observe that the enthalpy of products (H}_p\text{) is higher than the enthalpy of reactants (H}_r\text{).}$ $\text{2. This means that }\Delta\text{H = H}_p - \text{H}_r > 0\text{, indicating an endothermic reaction (heat is absorbed from the surroundings).}$ $\text{3. The diagram also shows }\Delta_f\text{H as the "net heat absorbed from surroundings," confirming that }\Delta\text{H is positive.}$ $\text{However, to determine spontaneity, we need to know both }\Delta\text{H and }\Delta\text{S values, along with the temperature:}$ $\text{- If }\Delta\text{H} > 0\text{ and }\Delta\text{S} > 0\text{, the reaction is spontaneous only at high temperatures (when T}\Delta\text{S} > \Delta\text{H).}$ $\text{- If }\Delta\text{H} > 0\text{ and }\Delta\text{S} < 0\text{, the reaction is never spontaneous.}$ $\text{- If }\Delta\text{H} < 0\text{ and }\Delta\text{S} > 0\text{, the reaction is always spontaneous.}$ $\text{- If }\Delta\text{H} < 0\text{ and }\Delta\text{S} < 0\text{, the reaction is spontaneous only at low temperatures (when |T}\Delta\text{S}| < |\Delta\text{H}|).}$ $\text{While the enthalpy diagram tells us that }\Delta\text{H} > 0\text{, it provides no information about the entropy change (}\Delta\text{S) for the reaction. Without knowing }\Delta\text{S, we cannot determine whether the reaction is spontaneous or non-spontaneous.}$ $\text{If }\Delta\text{S is positive and large enough, the reaction could be spontaneous at high temperatures despite having a positive }\Delta\text{H. If }\Delta\text{S is negative or not sufficiently positive, the reaction would be non-spontaneous.}$ $\text{Therefore, we cannot predict the spontaneity of the reaction solely from the enthalpy diagram provided.}$ $\text{The correct answer is option 3: Cannot predict spontaneity of the reaction from the graph given above.}$

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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
}