For the reaction, N_2O_4 rightleftharpoons 2NO_2$N_{2}O_{4} \rightleftharpoons 2NO_{2}$, graph is plotted as shown below. Identify correct statements.
A. Standard free energy change for the reaction is -5.40text kJ mol^-1$-5.40\text{ kJ mol}^{-1}$.
B. As Delta G^ominus$\Delta G^{\ominus}$ in graph is positive, N_2O_4$N_{2}O_{4}$ will not dissociate into NO_2$NO_{2}$ at all.
C. Reverse reaction will go to completion.
D. When 1 mole of N_2O_4$N_{2}O_{4}$ changes into equilibrium mixture, value of Delta G = -0.84text kJ mol^-1$\Delta G = -0.84\text{ kJ mol}^{-1}$.
E. When 2 mole of NO_2$NO_{2}$ changes into equilibrium mixture, Delta G$\Delta G$ for equilibrium mixture is -6.24text kJ mol^-1$-6.24\text{ kJ mol}^{-1}$.
A plot of Gibbs free energy against the fraction of N2O4 dissociated.
Choose the correct answer from the options given below :
A.textD and E only$\text{D and E only}$
B.textC and E only$\text{C and E only}$
C.textA and D only$\text{A and D only}$
D.textB and C only$\text{B and C only}$
Solution & Explanation
### Core Logic
Let's analyze the statements based on the given Gibbs free energy (G) vs. extent of reaction plot.
A plot of Gibbs free energy against the fraction of N2O4 dissociated.
A. Standard free energy change (Delta_r G^circ$\Delta_r G^{\circ}$) is the difference between standard free energies of pure products (point B) and pure reactants (point A): Delta_r G^circ = G_B^circ - G_A^circ$\Delta_r G^{\circ} = G_B^{\circ} - G_A^{\circ}$. Since B is higher than A, Delta_r G^circ$\Delta_r G^{\circ}$ is positive, not -5.40text kJ mol^-1$-5.40\text{ kJ mol}^{-1}$. Statement A is false.
B. Even if Delta_r G^circ$\Delta_r G^{\circ}$ is positive, the minimum of the curve (equilibrium state E) lies between the pure reactant and product states. Therefore, partial dissociation occurs to reach equilibrium. It is false to say it will not dissociate at all. Statement B is false.
C. The minimum E is not at fraction = 0 or 1, meaning an equilibrium mixture exists. Reverse reaction does not go to completion. Statement C is false.
D. From 1 mole of pure reactant (point A) to the equilibrium mixture (point E), the drop in Gibbs energy is 0.84text kJ mol^-1$0.84\text{ kJ mol}^{-1}$. Thus Delta G = -0.84text kJ mol^-1$\Delta G = -0.84\text{ kJ mol}^{-1}$ is correct. Statement D is true.
E. The difference from pure products (point B, equivalent to 2 moles NO_2$NO_2$) to equilibrium (E) is the entire vertical distance. The drop from B to A is 5.40$5.40$, and A to E is 0.84$0.84$. So drop from B to E is - (5.40 + 0.84) = -6.24text kJ mol^-1$- (5.40 + 0.84) = -6.24\text{ kJ mol}^{-1}$. Statement E is true.
### Step 1: Final Conclusion
Only statements D and E are correct.
### Pattern Recognition
The free energy of mixing creates a minimum 'dip' (Equilibrium) below both pure reactants and pure products, preventing either forward or reverse reactions from going absolutely to 100% completion.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Class 11 Chemistry: Equilibrium
A.Delta G text is negative for a spontaneous reaction$\Delta G \text{ is negative for a spontaneous reaction}$
B.Delta G text is positive for a spontaneous reaction$\Delta G \text{ is positive for a spontaneous reaction}$
C.Delta G text is zero for a reversible reaction$\Delta G \text{ is zero for a reversible reaction}$
D.Delta G text is positive for a non-spontaneous reaction$\Delta G \text{ is positive for a non-spontaneous reaction}$
Solution
### Core Logic
According to the second law of thermodynamics, at constant temperature and pressure, the change in Gibbs free energy (Delta G$\Delta G$) dictates the spontaneity of a process.
- If Delta G lt 0$\Delta G \lt 0$ (negative), the process is spontaneous.
- If Delta G gt 0$\Delta G \gt 0$ (positive), the process is non-spontaneous.
- If Delta G = 0$\Delta G = 0$, the system is in equilibrium (reversible process).
Therefore, the statement "Delta G$\Delta G$ is positive for a spontaneous reaction" is factually incorrect.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q83jee_main_2024_30_january_eveningHess's Law of Constant Heat Summation
### Related Formula
According to Hess's Law, the net enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which it can be divided.
### Core Logic
Let the given reactions be:
(1) 2mathrmFe_(s) + frac32mathrmO_2(g) rightarrow mathrmFe_2mathrmO_3(s), quad Delta H_1 = -822 \, mathrmkJ/mol$2\mathrm{Fe}_{(s)} + \frac{3}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{Fe}_2\mathrm{O}_{3(s)}, \quad \Delta H_1 = -822 \, \mathrm{kJ/mol}$
(2) mathrmC_(s) + frac12mathrmO_2(g) rightarrow mathrmCO_(g), quad Delta H_2 = -110 \, mathrmkJ/mol$\mathrm{C}_{(s)} + \frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}, \quad \Delta H_2 = -110 \, \mathrm{kJ/mol}$
Target Reaction (3):
3mathrmC_(s) + mathrmFe_2mathrmO_3(s) rightarrow 2mathrmFe_(s) + 3mathrmCO_(g), quad Delta H_3 = ?$$3\mathrm{C}_{(s)} + \mathrm{Fe}_2\mathrm{O}_{3(s)} \rightarrow 2\mathrm{Fe}_{(s)} + 3\mathrm{CO}_{(g)}, \quad \Delta H_3 = ?$$
To construct the target reaction:
- We need 3 mathrmCO_(g)$3 \mathrm{CO}_{(g)}$ on the product side, so we multiply reaction (2) by 3.
- We need mathrmFe_2mathrmO_3(s)$\mathrm{Fe}_2\mathrm{O}_{3(s)}$ on the reactant side and 2 mathrmFe_(s)$2 \mathrm{Fe}_{(s)}$ on the product side, so we reverse reaction (1).
### Step 1: Calculate Net Enthalpy
Target Reaction (3) = 3 times (2) - (1)$3 \times (2) - (1)$Delta H_3 = 3 times Delta H_2 - Delta H_1$$\Delta H_3 = 3 \times \Delta H_2 - \Delta H_1$$Delta H_3 = 3(-110) - (-822)$$\Delta H_3 = 3(-110) - (-822)$$Delta H_3 = -330 + 822 = 492 \, mathrmkJ/mol$$\Delta H_3 = -330 + 822 = 492 \, \mathrm{kJ/mol}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q83jee_main_2024_30_jan_morningWork Done in Cyclic Process
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path Arightarrow Brightarrow Crightarrow A$A\rightarrow B\rightarrow C\rightarrow A$ as shown in the diagram. The total work done in the process is ________ J.
The image is a graph of Volume (dm3) vs Pressure (kPa) showing a triangular cyclic process starting from A(10,10) to B(10,30) to C(30,10) and back to A.
Numerical Answer.Answer: 200 to 200
Solution
### Related Formula
W_textcyclic = textArea enclosed in P-V graph$$W_{\text{cyclic}} = \text{Area enclosed in P-V graph}$$
### Core Logic
The work done in a cyclic process is equal to the magnitude of the area enclosed by the cycle on a Pressure-Volume graph.
Note that the provided graph is Volume (V$V$) on the y-axis versus Pressure (P$P$) on the x-axis.
The path A rightarrow B rightarrow C rightarrow A$A \rightarrow B \rightarrow C \rightarrow A$ is traced in a clockwise direction on the V$V$-P$P$ graph.
Clockwise on a V$V$-P$P$ graph corresponds to anti-clockwise on a standard P$P$-V$V$ graph, meaning net expansion work is done by the gas, making it positive conventionally (or negative depending on chemistry sign convention, but magnitude is asked for).
### Step 1: Calculating Area
The enclosed region is a right-angled triangle.
Base of triangle on P-axis = 30 - 10 = 20 text kPa$30 - 10 = 20 \text{ kPa}$
Height of triangle on V-axis = 30 - 10 = 20 text dm^3$30 - 10 = 20 \text{ dm}^3$textArea = frac12 times textbase times textheight$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$textArea = frac12 times 20 times 20 = 200 text kPacdottextdm^3$$\text{Area} = \frac{1}{2} \times 20 \times 20 = 200 \text{ kPa}\cdot\text{dm}^3$$
### Step 2: Unit conversion
1 text kPa = 10^3 text Pa$1 \text{ kPa} = 10^3 \text{ Pa}$1 text dm^3 = 1 text Litre = 10^-3 text m^3$1 \text{ dm}^3 = 1 \text{ Litre} = 10^{-3} \text{ m}^3$W = 200 times 10^3 text Pa times 10^-3 text m^3$$W = 200 \times 10^3 \text{ Pa} \times 10^{-3} \text{ m}^3$$W = 200 text J$$W = 200 \text{ J}$$
### Pattern Recognition
1 text kPa cdot 1 text L = 1 text Joule$1 \text{ kPa} \cdot 1 \text{ L} = 1 \text{ Joule}$. This direct conversion saves time without converting explicitly to standard SI units (Pa$Pa$ and m^3$m^3$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q90jee_main_2024_31_jan_eveningWork Done in Isothermal Reversible Expansion
If 5text moles$5\text{ moles}$ of an ideal gas expands from 10text L$10\text{ L}$ to a volume of 100text L$100\text{ L}$ at 300text K$300\text{ K}$ under isothermal and reversible condition then work w$w$, is -xtext J$-x\text{ J}$. The value of x$x$ is ________
(Given R = 8.314text J K^-1textmol^-1$R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$)
Numerical Answer.Answer: 28720 to 28721
Solution
### Related Formula
W = -2.303 \, nRT log left( fracV_2V_1 right)$$W = -2.303 \, nRT \log \left( \frac{V_2}{V_1} \right)$$
### Core Logic
For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention.
Given:
n = 5text moles$n = 5\text{ moles}$R = 8.314text J K^-1textmol^-1$R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$T = 300text K$T = 300\text{ K}$V_1 = 10text L$V_1 = 10\text{ L}$V_2 = 100text L$V_2 = 100\text{ L}$
### Step 1: Calculating Work Done
W = -2.303 times 5 times 8.314 times 300 times logleft( frac10010 right)$$W = -2.303 \times 5 \times 8.314 \times 300 \times \log\left( \frac{100}{10} \right)$$W = -2.303 times 5 times 8.314 times 300 times log(10)$$W = -2.303 \times 5 \times 8.314 \times 300 \times \log(10)$$W = -2.303 times 12471 times 1$$W = -2.303 \times 12471 \times 1$$W = -28720.713text J$$W = -28720.713\text{ J}$$
### Step 2: Final Formatting
The question asks for work w = -xtext J$w = -x\text{ J}$.
So x = 28720.713$x = 28720.713$, which rounds to 28721$28721$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q88jee_main_2024_31_jan_morningGibbs Free Energy and Equilibrium
Consider the following reaction at 298 K.
frac32O_2(g) rightleftharpoons O_3(g). quad K_p = 2.47 times 10^-29$$\frac{3}{2}O_{2(g)} \rightleftharpoons O_{3(g)}. \quad K_p = 2.47 \times 10^{-29}$$Delta_rG^ominus$\Delta_rG^{\ominus}$ for the reaction is ________ kJ.
(Given R = 8.314 text J K^-1 mol^-1$R = 8.314 \text{ J } K^{-1} mol^{-1}$)
Numerical Answer.Answer: 163 to 164
Solution
### Related Formula
Delta_rG^ominus = -RT ln K_p$$\Delta_rG^{\ominus} = -RT \ln K_p$$
### Step 1: Calculation
Delta_rG^ominus = -8.314 times 10^-3 text kJ K^-1 mol^-1 times 298 text K times ln(2.47 times 10^-29)$$\Delta_rG^{\ominus} = -8.314 \times 10^{-3} \text{ kJ } K^{-1} mol^{-1} \times 298 \text{ K} \times \ln(2.47 \times 10^{-29})$$= -8.314 times 10^-3 times 298 times (-65.87)$$= -8.314 \times 10^{-3} \times 298 \times (-65.87)$$= 163.19 text kJ$$= 163.19 \text{ kJ}$$
### Step 2: Nearest Integer
Rounding 163.19$163.19$ to the nearest integer gives 163$163$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
More Thermodynamics Questions — jee_main_2026_21_jan_morning
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