Which of the following mixing of 1M base and 1M acid leads to the largest increase in temperature?

Solution & Explanation

### Related Formula Q = n_textreacted cdot Delta H_textneutralization Delta T = fracQm cdot c ### Core Logic The temperature rise depends directly on the total heat released (Q) normalized by the total heat capacity of the resulting mixed volume (m cdot c). Let's evaluate the millimoles of mathrmH^+ and mathrmOH^- that react in each mixture: 1. **Option 1:** 30text mL of 1mathrmM mathrmHCl + 30text mL of 1mathrmM mathrmNaOH textReactive millimoles = 30text mmol. Both are strong electrolytes, releasing full neutralization energy (sim -57.3text kJ/mol). Total volume = 60text mL. 2. **Option 2:** 30text mL of 1mathrmM mathrmCH_3COOH + 30text mL of 1mathrmM mathrmNaOH textReactive millimoles = 30text mmol. However, since acetic acid is a weak acid, part of the heat is consumed in its ionization. Thus, less total heat is evolved compared to Option 1. 3. **Option 3:** 50text mL of 1mathrmM mathrmHCl + 20text mL of 1mathrmM mathrmNaOH textLimiting reagent = mathrmNaOH = 20text mmol. Only 20text mmol reacts. Total volume = 70text mL. 4. **Option 4:** 45text mL of 1mathrmM mathrmCH_3COOH + 25text mL of 1mathrmM mathrmNaOH textLimiting reagent = 25text mmol weak neutralization profile. Comparing Option 1 and Option 3, Option 1 releases significantly more heat (30text mmol vs 20text mmol) into a smaller volume (60text mL vs 70text mL), yielding the largest increase in temperature Delta T. ### Pattern Recognition To maximize Delta T, look for the option that maximizes the amount of reacting strong acid and strong base equivalents while keeping the total solution volume as small as possible. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 7

Q78 jee_main_2024_29_jan_morning Spontaneity and Gibbs Free Energy
  • A. Delta G text is negative for a spontaneous reaction
  • B. Delta G text is positive for a spontaneous reaction
  • C. Delta G text is zero for a reversible reaction
  • D. 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) dictates the spontaneity of a process. - If Delta G lt 0 (negative), the process is spontaneous. - If Delta G gt 0 (positive), the process is non-spontaneous. - If Delta G = 0, the system is in equilibrium (reversible process). Therefore, the statement "Delta G is positive for a spontaneous reaction" is factually incorrect. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q83 jee_main_2024_30_january_evening Hess's Law of Constant Heat Summation
Two reactions are given below: 2mathrmFe_(s) + frac32mathrmO_2(g) rightarrow mathrmFe_2mathrmO_3(s), Delta mathrmH^circ = -822 mathrmkJ/mol mathrmC_(s) + frac12mathrmO_2(g) rightarrow mathrmCO_(g), Delta mathrmH^circ = -110 mathrmkJ/mol Then enthalpy change for following reaction 3mathrmC_(s) + mathrmFe_2mathrmO_3(s) rightarrow 2mathrmFe_(s) + 3mathrmCO_(g)
Numerical Answer. Answer: 492 to 492

Solution

### 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) mathrmC_(s) + frac12mathrmO_2(g) rightarrow mathrmCO_(g), quad Delta H_2 = -110 \, mathrmkJ/mol Target Reaction (3): 3mathrmC_(s) + mathrmFe_2mathrmO_3(s) rightarrow 2mathrmFe_(s) + 3mathrmCO_(g), quad Delta H_3 = ? To construct the target reaction: - We need 3 mathrmCO_(g) on the product side, so we multiply reaction (2) by 3. - We need mathrmFe_2mathrmO_3(s) on the reactant side and 2 mathrmFe_(s) on the product side, so we reverse reaction (1). ### Step 1: Calculate Net Enthalpy Target Reaction (3) = 3 times (2) - (1) Delta H_3 = 3 times Delta H_2 - Delta H_1 Delta H_3 = 3(-110) - (-822) Delta H_3 = -330 + 822 = 492 \, mathrmkJ/mol ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q83 jee_main_2024_30_jan_morning Work 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 as shown in the diagram. The total work done in the process is ________ J.
Work Done in Cyclic Process diagram for Q83 - JEE Main 2024 Morning
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 ### 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) on the y-axis versus Pressure (P) on the x-axis. The path A rightarrow B rightarrow C rightarrow A is traced in a clockwise direction on the V-P graph. Clockwise on a V-P graph corresponds to anti-clockwise on a standard P-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 Height of triangle on V-axis = 30 - 10 = 20 text dm^3 textArea = frac12 times textbase times textheight textArea = frac12 times 20 times 20 = 200 text kPacdottextdm^3 ### Step 2: Unit conversion 1 text kPa = 10^3 text Pa 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 text J ### Pattern Recognition 1 text kPa cdot 1 text L = 1 text Joule. This direct conversion saves time without converting explicitly to standard SI units (Pa and m^3). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q90 jee_main_2024_31_jan_evening Work Done in Isothermal Reversible Expansion
If 5text moles of an ideal gas expands from 10text L to a volume of 100text L at 300text K under isothermal and reversible condition then work w, is -xtext J. The value of x is ________ (Given R = 8.314text J K^-1textmol^-1)
Numerical Answer. Answer: 28720 to 28721

Solution

### Related Formula W = -2.303 \, nRT log left( fracV_2V_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 R = 8.314text J K^-1textmol^-1 T = 300text K V_1 = 10text L V_2 = 100text 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(10) W = -2.303 times 12471 times 1 W = -28720.713text J ### Step 2: Final Formatting The question asks for work w = -xtext J. So x = 28720.713, which rounds to 28721. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q88 jee_main_2024_31_jan_morning Gibbs 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 Delta_rG^ominus for the reaction is ________ kJ. (Given 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 ### 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) = -8.314 times 10^-3 times 298 times (-65.87) = 163.19 text kJ ### Step 2: Nearest Integer Rounding 163.19 to the nearest integer gives 163. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

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