Question 1

Consider the reaction 2 H _2 S (g) + 3 O _2(g) 2 H _2 O (l) + 2 SO _2(g) The magnitude of enthalpy change for the reaction in kJ mol^ -1 is ________. (Nearest integer) Given: _fH^ ( H _2 S ) = -20.1 k

Accepted Answer

The standard enthalpy of reaction is given by: _r H^ = _f H^ ( products ) - _f H^ ( reactants ) For the given reaction: 2 H _2 S (g) + 3 O _2(g) 2 H _2 O (l) + 2 SO _2(g) _r H^ = [2 _f H^ ( H _2 O , l) + 2 _f H^ ( SO _2, g)] - [2 _f H^ ( H _2 S , g) + 3 _f H^ ( O _2, g)] Substituting the given values: _r H^ = [2(-286.0) + 2(-297.0)] - [2(-20.1) + 3(0)] _r H^ = [-572.0 - 594.0] - [-40.2] _r H^ = -1166.0 + 40.2 = -1125.8 kJ mol ^ -1 The magnitude of the enthalpy change is 1125.8 kJ mol ^ -1 . Roun

Question 2

Match List - I with List - II. Given V_1 and V_2 are initial and final volumes respectively. List - I (Isothermal process) List - II (Expression) A. Reversible expansion I. q = 0 B. Free expansion II.

Accepted Answer

For an isothermal reversible expansion, the change in internal energy U = 0. From the first law of thermodynamics, q = -w. The work done is w = -nRT ( V_2 V_1 ), which gives q = nRT ( V_2 V_1 ). Thus, A matches II. For free expansion, the external pressure is zero (p_ ext = 0), so the work done w = 0. Since the process is isothermal, U = 0. Therefore, the heat exchanged q = 0. Thus, B matches I. For an irreversible compression, the work done against a constant external pressure is given by w = -

Question 3

Arrange the following isothermal processes in order of the magnitude of the work (p - V) involved between states 1 and 2. A. Expansion in single stage w_A B. Expansion in multi stages w_B C. Compressi

Accepted Answer

Let State 1 have pressure P_1 and volume V_1, and State 2 have pressure P_2 and volume V_2. For an isothermal expansion, V_2 > V_1 and P_1 > P_2. For expansion from State 1 to State 2: In a single-stage expansion, the external pressure is constant at the final pressure P_2. The magnitude of work is |w_A| = P_2(V_2 - V_1). In a multi-stage expansion, the external pressure is reduced in steps, keeping it closer to the internal pressure of the gas. The work done by the gas increases with the number

Question 4

Consider the reaction X Y at 300 K. If H^ and K are 28.40 kJ mol^ -1 and 1.8 10^ -7 at the same temperature, then the magnitude of S^ for the reaction in J K^ -1 mol^ -1 is _______. (Nearest integer)

Accepted Answer

The standard Gibbs free energy change is given by: G^ = -RT K = -2.3 RT K First, calculate K: K = (1.8 10^ -7 ) = (18 10^ -8 ) K = 18 - 8 = (2 3^2) - 8 K = 2 + 2 3 - 8 Substituting the given values: K = 0.30 + 2(0.48) - 8 = 0.30 + 0.96 - 8 = 1.26 - 8 = -6.74 Now, calculate G^ : G^ = -2.3 8.3 300 (-6.74) = 38599.98 J mol ^ -1 Using the relation G^ = H^ - T S^ : 38599.98 = 28400 - 300 S^ 300 S^ = 28400 - 38599.98 = -10199.98 S^ = - 10199.98 300 -34 J K ^ -1 mol ^ -1 The magnitude of S^ is 34. Answ

Question 5

Consider the following data for the reaction X_2(g) + Y_2(g) 2XY(g) at 600 K . The _r G^ (in kJ mol^ -1 ) for the reaction is : Compound _f H^ _ 600K (kJ mol^ -1 ) S^ _ 600K (J mol^ -1 K^ -1 ) XY(g) 4

Accepted Answer

The given reaction is X_2(g) + Y_2(g) 2XY(g). The standard enthalpy of reaction, _r H^ , is calculated as: _r H^ = _f H^ ( products ) - _f H^ ( reactants ) _r H^ = 2 _f H^ (XY) - [ _f H^ (X_2) + _f H^ (Y_2)] _r H^ = 2(42) - (8 + 80) = 84 - 88 = -4 kJ mol ^ -1 The standard entropy of reaction, _r S^ , is calculated as: _r S^ = S^ ( products ) - S^ ( reactants ) _r S^ = 2 S^ (XY) - [S^ (X_2) + S^ (Y_2)] _r S^ = 2(200) - (140 + 250) = 400 - 390 = 10 J K ^ -1 mol ^ -1 Converting _r S^ to kJ K ^ -1 m

Question 6

The correct order of molar heat capacities measured at 298 K and 1 bar is :

Accepted Answer

The molar heat capacity of Helium (a monatomic gas) at constant pressure is C_p = 5 2 R 20.8 J K ^ -1 mol ^ -1 . The molar heat capacity of Copper (a solid) follows the Dulong-Petit law, which gives C_p 3R 24.9 J K ^ -1 mol ^ -1 . The molar heat capacity of Bromine (a diatomic liquid) is significantly higher due to translational, rotational, and vibrational degrees of freedom along with intermolecular interactions in the liquid state, having C_p 75.7 J K ^ -1 mol ^ -1 . Therefore, the correct or

Question 7

If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The | H_f°| of ethanol at 298.15 K is ______ 10^2 kJ mol^ -1 . (Nearest integer) Given:

Accepted Answer

Molar mass of ethanol (C_2H_5OH) = 2(12) + 6(1) + 16 = 46 g mol^ -1 Number of moles of ethanol burnt, n = 3.365 46 = 0.07315 mol Since the combustion occurs in a bomb calorimeter (constant volume), the heat produced corresponds to the change in internal energy ( U_c^ ). U_c^ = - 99.472 0.07315 = -1359.8 kJ mol^ -1 The balanced chemical equation for the combustion of ethanol is: C_2H_5OH(l) + 3O_2(g) 2CO_2(g) + 3H_2O(l) Change in the number of gaseous moles, n_g = 2 - 3 = -1 Now, calculating the

Question 8

Given below are two statements: Statement I: For an ideal gas, heat capacity at constant volume is always greater than the heat capacity at constant pressure. Statement II: In a constant volume proces

Accepted Answer

For an ideal gas, the relationship between heat capacity at constant pressure (C_p) and heat capacity at constant volume (C_v) is given by Mayer's relation: C_p - C_v = R Since the universal gas constant R is a positive value, C_p > C_v. Thus, Statement I is false. In a constant volume (isochoric) process, the change in volume V = 0. Work done, W = P V = 0. According to the first law of thermodynamics: U = q + W Since W = 0, we get U = q_v. This means that in a constant volume process, no work i

Question 9

Gas 'A' undergoes change from state 'X' to state 'Y'. In this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state 'X' by another process

Accepted Answer

Using the first law of thermodynamics, U = q + w. For the process from state X to state Y: Heat absorbed by the gas, q_ X Y = +10 J Work done by the gas, w_ X Y = -18 J Change in internal energy, U_ X Y = q_ X Y + w_ X Y = 10 - 18 = -8 J For the reverse process from state Y to state X: Change in internal energy, U_ Y X = - U_ X Y = +8 J Heat evolved by the gas, q_ Y X = -6 J Let the work done on the gas be w_ Y X . U_ Y X = q_ Y X + w_ Y X 8 = -6 + w_ Y X w_ Y X = +14 J Since the work done is po

Question 10

Consider the following data. (i) 2Al(s) + 6HCl(aq) Al_2Cl_6(aq) + 3H_2(g) + 1200 kJ/mol (ii) H_2(g) + Cl_2(g) 2HCl(g) + 164 kJ/mol (iii) HCl(g) + aq HCl(aq) + 83 kJ/mol (iv) Al_2Cl_6(s) + aq Al_2Cl_6(

Accepted Answer

The given reactions with their enthalpy changes are: (i) 2Al(s) + 6HCl(aq) Al_2Cl_6(aq) + 3H_2(g), H_1 = -1200 kJ/mol (ii) H_2(g) + Cl_2(g) 2HCl(g), H_2 = -164 kJ/mol (iii) HCl(g) + aq HCl(aq), H_3 = -83 kJ/mol (iv) Al_2Cl_6(s) + aq Al_2Cl_6(aq), H_4 = -663 kJ/mol The required reaction for the enthalpy of formation of anhydrous solid Al_2Cl_6 is: 2Al(s) + 3Cl_2(g) Al_2Cl_6(s) This target equation can be obtained by applying the following algebraic operations on the given equations: Equation (i)

Question 11

At the transition temperature T, A B and G^0 = 105 - 35 T where A and B are two states of substance X. The transition temperature in °C when pressure is 1 atm is _______. (Nearest integer)

Accepted Answer

At the transition temperature and 1 atm pressure, the two states A and B are in equilibrium. For a phase transition of a pure substance at 1 atm, both states are in their standard states, so the reaction quotient Q = 1. Using the relation G = G^0 + RT Q, we get G = G^0. At equilibrium, G = 0, which implies G^0 = 0. Given G^0 = 105 - 35 T, we can equate it to zero: 105 - 35 T = 0 35 T = 105 T = 3 T = 10^3 = 1000 K Converting the temperature to Celsius: T(^ C ) = 1000 - 273 = 727^ C Answer: 727

Question 12

The plot of _ 10 ~K vs 1 ~T gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).

Accepted Answer

The relationship between the standard Gibbs free energy change and the equilibrium constant is given by G^o = -RT K. We also know that G^o = H^o - T S^o. Equating the two expressions: -RT K = H^o - T S^o. Dividing by -RT, we get: K = - H^o RT + S^o R . To convert natural log to base 10, use K = 2.303 _ 10 K: 2.303 _ 10 K = - H^o RT + S^o R . Dividing by 2.303: _ 10 K = - H^o 2.303RT + S^o 2.303R . Rearranging in the form of a straight line equation y = mx + c where y = _ 10 K and x = 1 T : _ 10

Question 13

20.0 dm ^ 3 of an ideal gas ' X ' at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure of the gas is 0.2 MPa. Which of the following option is correct? (Given: 2=0.3010 and 5=

Accepted Answer

For isothermal reversible process U = H = 0 w_ iso = -p_1 v_1 P_1 P_2 w_ iso = -0.5 10^6 20 10^ -3 0.5 0.2 w_ iso = -0.5 10^6 20 10^ -3 2.303 (.6989 - .3010) w -9.1 kJ q = -w = 9.1 kJ

Question 14

The heat of atomisation of methane and ethane are ' x ' kJ mol ^ -1 and ' y ' kJ mol ^ -1 respectively. The longest wavelength ( ) of light capable of breaking the C - C bond can be expressed in SI un

Accepted Answer

Heat of atomization: CH₄ has x kJ/mol (4 C-H bonds), C₂H₆ has y kJ/mol (1 C-C + 6 C-H bonds) Let C-H bond energy = a, C-C bond energy = b Then: x = 4a and y = b + 6a Solving: a = x/4, so b = y - 6x/4 = (4y - 6x)/4 kJ/mol Converting to J/mol: b = (4y - 6x) × 250 J/mol Using E = hc and = hc E : = N_A hc 250(4y - 6x) (in SI units)

Question 15

Match the LIST-I with LIST-II \( array |l|l|c|l| List-I & Isothermal process \ & for ideal gas system & List-II & Work done (V_f > V_i) \ A. & Reversible expansion & I. & w = 0 \ B. & Free expansio

Accepted Answer

Matching isothermal processes with work expressions: A. Reversible expansion: In a reversible isothermal expansion from V_i to V_f (where V_f > V_i), the gas does work against external pressure equal to the surroundings. The work is w = nRT V_f V_i , matching II. B. Free expansion: The gas expands into a vacuum with no external pressure opposition. No work is done: w = 0, matching I. C. Irreversible expansion: Isothermal expansion from V_i to V_f against constant external pressure p_ ex . Work d

Question 16

A D is an endothermic reaction occurring in three steps (elementary). (i) A B H _ i =+ ve (ii) B C H _ i i =- ve (iii) C D H _ i i i =- ve Which of the following graphs between potential energy (y-axi

Accepted Answer

The reaction A → D occurs through three elementary steps with energy changes: Step (i): A → B is endothermic ( H_I = +ve), so B has higher energy than A, creating the first peak. Step (ii): B → C is exothermic ( H_ II = -ve), so C has lower energy than B, creating a lower second peak. Step (iii): C → D is exothermic ( H_ III = -ve), so D has lower energy than C, creating an even lower third feature. Overall reaction: Since step (i) is endothermic and steps (ii) and (iii) are exothermic, the reac

Question 17

A cup of water at 5^ C (system) is placed in a microwave oven and the oven is turned on for one minute during which the water begins to boil. Which of the following option is true ?

Accepted Answer

For water heating in a microwave: Heat (q): The microwave radiates electromagnetic energy that is absorbed by water molecules, so q = +ve Work (w): As water heats and boils, it expands against atmospheric pressure. The system does work on the surroundings, so w = -ve Internal Energy (ΔU): From the first law, ΔU = q + w. Since heat absorbed (+ve) is much larger in magnitude than work done against atmosphere (-ve), the net change is ΔU = +ve (confirmed by temperature increase from 5°C to boiling)

Question 18

If the enthalpy of sublimation of Li is 155 ~kJ ~mol ^ -1 , enthalpy of dissociation of F _ 2 is 150 ~kJ ~mol ^ -1 , ionization enthalpy of Li is 520 ~kJ ~mol ^ -1 , electron gain enthalpy of F is -31

Accepted Answer

Using Born-Haber cycle: H_f(LiF) = H_ sub (Li) + 1 2 H_ diss (F_2) + H_ ie (Li) + H_ eg (F) - U_ lattice Substituting values: -594 = 155 + 1 2 (150) + 520 + (-313) - U_ lattice -594 = 155 + 75 + 520 - 313 - U_ lattice -594 = 437 - U_ lattice U_ lattice = 437 + 594 = 1031 kJ/mol

Question 19

Match the LIST-I with LIST-II \( array |c|l|c|c| & List-I (Thermodynamic Process) & & List-II \ & & & (Magnitude in kJ) \ A. & array l Work done in reversible, \ isothermal expansion of \ 2 mol id

Accepted Answer

Calculate each thermodynamic quantity: A. Reversible isothermal expansion (2 mol, 2→20 dm³, 300 K): W = nRT (V_2/V_1) = 2 8.314 300 (10) = 2 8.314 300 2.303 = 11,481 J ≈ 11.5 kJ → II B. Irreversible isothermal expansion (1 mol, 1→3 m³, 3 kPa constant pressure): W = P_ ext V = 3000 (3-1) = 6000 J = 6 kJ → III C. Adiabatic expansion (1 mol, T = -320 K, C_V = 3 2 R): U = nC_V| T| = 1 3 2 8.314 320 = 3,991 J ≈ 4 kJ → I D. Enthalpy change (1 mol, T = 337 K, C_p = 5 2 R): H = nC_p T = 1 5 2 8.314 337

Question 20

Consider the following data: _ f H ^ (methane, g)=- X kJ mol ^ -1 Enthalpy of sublimation of graphite = Y kJ mol ^ -1 Dissociation enthalpy of H _ 2 = Z kJ mol ^ -1 The bond enthalpy of C - H bond is

Accepted Answer

For methane formation: C(g) + 2 H _2(g) CH _4(g). Using enthalpy balance: _f H° = Bonds broken - Bonds formed . Breaking: sublimation of C (Y) and 2 H-H bonds (2Z); Forming: 4 C-H bonds. Energy equation: -X = Y + 2Z - 4B where B is C-H bond enthalpy. Solving: 4B = X + Y + 2Z, so B = X + Y + 2Z 4 .

Thermodynamics (C) — Previous Year Questions & Solutions