Question

For the reaction, \(N_{2}O_{4} \rightleftharpoons 2NO_{2}\) graph is plotted as shown below. Identify correct statements.
A. Standard free energy change for the reaction is 5.40 kJ \(mol^{-1}\). 
B. As \(\Delta G\) in graph is positive, \(N_{2}O_{4}\) will not dissociate into \(NO_{2}\) at all. 
C. Reverse reaction will go to completion. 
D. When 1 mole of \(N_{2}O_{4}\) changes into equilibrium mixture, value of \(\Delta G = -0.84 \text{ kJ mol}^{-1}\). 
E. When 2 mole of \(NO_{2}\) changes into equilibrium mixture, \(\Delta G\) for equilibrium mixture is \(-6.24 \text{ kJ mol}^{-1}\). 

Choose the correct answer from the following.

• A. B and C only
• B. A and D only
• C. D and E only
• D. C and E only

Hint:

\(\Delta G^\circ\) determines the position of equilibrium (Ratio of Products/Reactants at minimum G), but \(\Delta G\) determines the spontaneity of moving towards equilibrium from a specific state.

Answer 1

The Correct Option is B

   Step 1: Analyze the Graph and Data
The graph plots Gibbs Free Energy (\(G\)) vs. Fraction of \(N_2O_4\) dissociated. \begin{itemize} \item Point A (Pure \(N_2O_4\)) corresponds to \(\xi = 0\). \item Point B (Pure \(2NO_2\)) corresponds to \(\xi = 1\). \item The minimum of the curve represents the equilibrium position. \end{itemize} From the graph labels (implied by typical context of this question): \begin{itemize} \item The drop in \(G\) from Pure Reactants to Equilibrium is \(0.84 \text{ kJ mol}^{-1}\). (i.e., \(G_{eq} - G_{reactants} = -0.84\)). \item The drop in \(G\) from Pure Products to Equilibrium is \(6.24 \text{ kJ mol}^{-1}\). (i.e., \(G_{eq} - G_{products} = -6.24\)). \end{itemize} Step 2: Evaluate Statement A
Standard Free Energy Change (\(\Delta G^\circ\)) is the difference between the standard G of products and reactants. \[ \Delta G^\circ = G^\circ(\text{Pure Products}) - G^\circ(\text{Pure Reactants}) \] Using the equilibrium drops: \[ G_{products} = G_{eq} + 6.24 \] \[ G_{reactants} = G_{eq} + 0.84 \] \[ \Delta G^\circ = (G_{eq} + 6.24) - (G_{eq} + 0.84) = 6.24 - 0.84 = +5.40 \text{ kJ mol}^{-1} \] Statement A is Correct. Step 3: Evaluate Other Statements
\begin{itemize} \item B & C: Since equilibrium is reached at a minimum \(G\) (intermediate composition), the reaction does not go to completion in either direction, nor does it fail to react. Statements B and C are Incorrect. \item D: This refers to the free energy change when starting from 1 mole of reactants to reach equilibrium. \(\Delta G = -0.84 \text{ kJ}\). Statement D is Correct. \item E: This refers to the free energy change when starting from products (2 moles of \(NO_2\)) to reach equilibrium. \(\Delta G = -6.24 \text{ kJ}\). Statement E is also physically Correct. \end{itemize} Step 4: Select Option
Since A and D are definitely correct and form Option 2, this is the intended answer.