Question 1

Using the following thermochemical equations, determine the enthalpy of vaporisation of water:
H₂(g) + 1/2 O₂(g) → H₂O(l), ΔH = -286 kJ
H₂(g) + 1/2 O₂(g) → H₂O(g), ΔH = -245.5 kJ

Accepted Answer

40.5 kJ. H2O(l) -> H2O(g): dHvap = (-245.5) - (-286) = 40.5 kJ. The stored 62.3 kJ is wrong.

Question 2

Given that the standard enthalpy of formation of NH₃ is -46.0 kJ mol⁻¹, and the enthalpy of formation of H₂ from gaseous atoms is -436 kJ mol⁻¹ while that of N₂ from gaseous atoms is -712 kJ mol⁻¹, what is the average N–H bond enthalpy in NH₃?

Accepted Answer

+352 kJ mol⁻¹. The average N–H bond enthalpy in NH₃ can be calculated by considering the enthalpy changes associated with the formation of NH₃ from its elements. The positive value indicates that energy is required to break the N–H bonds, which is consistent with bond enthalpy being a measure of bond strength.

Question 3

For the spontaneous reaction
2C₂H₆(g) + 25O₂(g) → 16CO₂(g) + 18H₂O(g),
what are the signs of ΔH, ΔS and ΔG in that order?

Accepted Answer

-, +, -. The reaction is exothermic, releasing heat, which gives it a negative ΔH. The increase in the number of gas molecules from reactants to products indicates an increase in entropy, resulting in a positive ΔS. Since the reaction is spontaneous at standard conditions, ΔG is negative.

Question 4

The enthalpy changes for these reactions are given:

1/2 H2(g) + 1/2 O2(g) → OH(g); ΔH = 42.09 kJ mol⁻¹
H2(g) → 2H(g); ΔH = 435.89 kJ mol⁻¹
O2(g) → 2O(g); ΔH = 495.05 kJ mol⁻¹

Using this data, determine the O–H bond energy in the hydroxyl radical.

Accepted Answer

423.38 kJ mol⁻¹. The O–H bond energy in the hydroxyl radical is calculated by considering the enthalpy changes of the given reactions. By applying Hess's law and combining the enthalpy values appropriately, we find that the bond energy corresponds to the correct option of 423.38 kJ mol⁻¹.

Question 5

The enthalpy of combustion of carbon to form carbon dioxide is −393.5 kJ per mole. What is the heat evolved when 35.2 g of CO₂ is produced from carbon and oxygen gas?

Accepted Answer

−315 kJ. The enthalpy of combustion indicates that for every mole of CO₂ produced, 393.5 kJ of heat is released. Since 35.2 g of CO₂ corresponds to 0.8 moles, the total heat evolved is 0.8 moles multiplied by -393.5 kJ/mole, resulting in -315 kJ.

Question 6

For a reaction with ΔH = 35.5 kJ mol⁻¹ and ΔS = 83.6 J K⁻¹ mol⁻¹, at what temperature condition will the reaction be spontaneous? (Assume ΔH and ΔS remain constant with temperature.)

Accepted Answer

T > 425 K. The reaction will be spontaneous when the Gibbs free energy change (ΔG) is negative, which occurs when the temperature is high enough to overcome the positive enthalpy change (ΔH) by utilizing the positive entropy change (ΔS). By calculating the temperature at which ΔG equals zero, we find that it is above 425 K, making option A the correct choice.

Question 7

An endothermic process is not spontaneous at the freezing point of water but turns spontaneous at the boiling point of water. Which statement about its thermodynamic quantities is correct?

Accepted Answer

Both ΔH and ΔS are positive.. In an endothermic process, heat is absorbed, which means ΔH is positive. Additionally, for the process to become spontaneous at higher temperatures, the entropy change ΔS must also be positive, indicating an increase in disorder.

Question 8

At 25°C, the standard enthalpies of formation of CO₂(g), H₂O(l), and glucose(s) are −400 kJ/mol, −300 kJ/mol, and −1300 kJ/mol, respectively. What is the standard enthalpy of combustion per gram of glucose at 25°C?

Accepted Answer

−16.11 kJ. The standard enthalpy of combustion of glucose is calculated by taking the enthalpy of formation of the products (CO₂ and H₂O) and subtracting the enthalpy of formation of glucose. This results in a negative value, indicating that combustion is an exothermic reaction, and when divided by the molar mass of glucose, it yields approximately −16.11 kJ per gram.

Question 9

The standard molar entropies of X₂, Y₂, and X₃Y₂ are 60, 40, and 50 J K⁻¹ mol⁻¹, respectively. For the reaction
1/2 X₂ + 3/2 Y₂ → X₃Y₂,
if ΔH = −30 kJ, the temperature at which the system is in equilibrium is

Accepted Answer

750 K. The equilibrium temperature can be determined using the Gibbs free energy equation, which relates enthalpy, entropy, and temperature. In this case, the negative enthalpy change and the calculated change in entropy for the reaction indicate that the reaction is spontaneous at 750 K, making it the correct answer.

Question 10

The bond enthalpies of H₂, X₂ and HX are given in the ratio 2: 1: 2. If the enthalpy of formation of HX is −50 kJ mol⁻¹, what is the bond enthalpy of X₂?

Accepted Answer

100 kJ mol⁻¹. With D(H2):D(X2):D(HX) = 2:1:2 = 2a:a:2a, dHf(HX) = a + a/2 - 2a = -a/2 = -50, so a = 100. The X2 bond enthalpy is 100 kJ/mol, not 200.

Question 11

At 298 K, consider the equilibrium reaction: 2NO(g) + O2(g) ⇌ 2NO2(g). If the standard free energy of formation of NO(g) is 86.6 kJ mol⁻¹ at 298 K and the equilibrium constant Kp is 1.6 × 10¹², what is the standard free energy of formation of NO2(g) at 298 K?

Accepted Answer

0.5[2 × 86600 − R(298) ln(1.6 × 10¹²)]. This option correctly applies the relationship between the standard free energies of formation and the equilibrium constant, taking into account the stoichiometry of the reaction. It averages the contributions from both reactants and products, leading to the correct calculation for the standard free energy of formation of NO2.

Question 12

A fuel cell operates with methanol as the fuel and oxygen as the oxidant. The overall reaction is
CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l)
At 298 K, the standard Gibbs energies of formation of CH3OH(l), H2O(l), and CO2(g) are −166.2, −237.2, and −394.4 kJ mol−1, respectively. If the standar

Accepted Answer

97%. delta G = [-394.4 + 2(-237.2)] - (-166.2) = -702.6 kJ. Efficiency = delta G / delta H = 702.6/726 = 0.968, i.e. about 97%.

Question 13

For a reversible reaction, if the activation energy of the forward process is 50 kcal, what can be said about the activation energy of the reverse process?

Accepted Answer

It may be either greater than or less than 50 kcal. The activation energy of the reverse process can vary depending on the energy difference between the reactants and products; thus, it can be either greater than or less than the activation energy of the forward process.

Question 14

Which thermodynamic condition is favorable for converting a metal oxide into the metal?

Accepted Answer

ΔH is negative and TΔS is positive at any temperature. Conversion is favorable when dG = dH - TdS < 0. If dH is negative and TdS is positive, dG is negative at all temperatures, so this condition is favorable. (dH positive with TdS negative makes dG always positive - never favorable.)

Question 15

In an Ellingham diagram, the line for which reaction has a negative slope with increasing temperature?

Accepted Answer

C + 1/2 O2 → CO. The reaction C + 1/2 O2 → CO has a negative slope in an Ellingham diagram because it represents the formation of carbon monoxide from carbon and oxygen, which becomes more favorable at higher temperatures, indicating that the entropy change is positive and the Gibbs free energy decreases.

Question 16

Consider these two reactions:
A + B → C + D, with ΔG° = +x kJ
D + E → F, with ΔG° = −y kJ
For the overall reaction A + B + E → C + F to be spontaneous, which statement is correct?

Accepted Answer

x < y. Adding the two steps, the overall dG = (+x) + (-y) = x - y. For spontaneity the overall dG must be negative, so x - y < 0, i.e. x < y.

Question 17

A closed adiabatic container fitted with a frictionless movable piston contains 1 mol of gaseous X at 300 K. The external pressure is kept fixed at 1 atm. A reaction is initiated using a negligible amount of electrical energy: 3X(g) → 2Y(g), with ΔH = −30 kJ mol⁻¹. The process proceeds at

Accepted Answer

600 K. The reaction releases heat (exothermic) and produces fewer moles of gas, which increases the temperature of the system. Given the heat released and the specific heat capacities, the final temperature can be calculated, resulting in 600 K after accounting for the heat transfer and the change in moles.

Question 18

An endothermic process is not spontaneous at the freezing point of water but turns spontaneous at the boiling point of water. Which statement about its thermodynamic quantities is correct?

Accepted Answer

Both ΔH and ΔS are positive. For an endothermic process delta H is positive. It is non-spontaneous at low T but spontaneous at high T, which from delta G = delta H - T*delta S requires delta S to be positive. Hence both delta H and delta S are positive.

Question 19

A heat engine takes in an amount of heat Q1 at temperature T1 and an amount of heat Q2 at temperature T2. If the work output of the engine is J(Q1 + Q2), then this situation

Accepted Answer

does not break the first law of thermodynamics. If the work output equals the net heat input (Q1+Q2), energy is conserved and the first law of thermodynamics is satisfied; such a process does not break the first law.

Question 20

Consider the following reactions:
2C + O2 → 2CO2; ΔH = −0393 J
2Zn + O2 → 2ZnO; ΔH = −412 J
Which statement is supported by these reactions?

Accepted Answer

Carbon is able to oxidise zinc. The reactions show that carbon can react with oxygen to form carbon dioxide, releasing energy, while zinc also reacts with oxygen to form zinc oxide. Since both reactions are exothermic, and carbon is capable of oxidizing zinc, it indicates that carbon can indeed oxidize zinc.

Question 21

What is the term for the amount of heat needed to increase the temperature of a body by 1 K?

Accepted Answer

heat capacity. Heat capacity is defined as the amount of heat required to raise the temperature of a body by 1 K (or 1 degree C). Specific heat is per unit mass, so for the whole body the correct term is heat capacity.

Question 22

The change in internal energy for a system going from state A to state B is 40 kJ/mol. If the system reaches B from A through a reversible process and then comes back to A through an irreversible process, what is the overall change in internal energy for the complete cycle?

Accepted Answer

Zero. The overall change in internal energy for a complete cycle is always zero, regardless of the path taken, because internal energy is a state function that depends only on the initial and final states.

Question 23

At 298 K, the bond enthalpies of C—H, C—C, C=C and H—H bonds are 414, 347, 615 and 435 kJ mol−1, respectively. What is the enthalpy change for the reaction
H2C=CH2(g) + H2(g) → H3C—CH3(g) at 298 K?

Accepted Answer

−125 kJ. Broken: C=C + 4 C-H + H-H = 615 + 4(414) + 435 = 2706 kJ. Formed: C-C + 6 C-H = 347 + 6(414) = 2831 kJ. dH = 2706 - 2831 = -125 kJ.

Question 24

For an irreversible change occurring at constant temperature and pressure, with only pressure–volume work involved, which relation between the entropy change and Gibbs free energy change is correct?

Accepted Answer

(dS) at constant V and E is positive, and (dG) at constant T and P is negative. Criteria for a spontaneous (irreversible) process: entropy of the isolated system increases, so (dS) at constant U,V is positive, while (dG) at constant T,P is negative. dS = 0 corresponds only to reversible/equilibrium conditions.

Question 25

Which expression correctly relates the Gibbs free energy change for a reaction to its equilibrium constant Kc?

Accepted Answer

−ΔG° = RT ln Kc. The correct thermodynamic relation uses the standard free energy: delta G(standard) = -RT ln Kc, equivalently -delta G(standard) = RT ln Kc.

Question 26

The enthalpy change of a reaction is independent of which of the following?

Accepted Answer

the sequence of intermediate steps in the reaction pathway. The enthalpy change of a reaction is determined solely by the initial and final states of the reactants and products, making it independent of the specific pathway or sequence of steps taken to achieve that change.

Question 27

An ideal gas expands in volume from 1 × 10⁻³ to 1 × 10⁻² m³ at 300 K against a constant pressure of 1 × 10⁵ N m⁻². The work done is

Accepted Answer

−900 J. Work done on the gas during expansion against constant external pressure = -P_ext*Delta_V = -(1e5)(1e-2 - 1e-3) = -900 J.

Question 28

The enthalpies of combustion of carbon and carbon monoxide are −393.5 and −283 kJ mol⁻¹ respectively. The enthalpy of formation of carbon monoxide per mole is

Accepted Answer

−110.5 kJ. C + 1/2 O2 -> CO equals (C + O2 -> CO2) minus (CO + 1/2 O2 -> CO2): dHf = -393.5 - (-283) = -110.5 kJ/mol.

Question 29

Consider the reaction: N2 + 3H2 → 2NH3, carried out at constant temperature and pressure. If ΔH and ΔU are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

Accepted Answer

ΔH < ΔU. For N2 + 3H2 -> 2NH3, delta n_gas = 2 - 4 = -2. Since delta H = delta U + (delta n_gas)RT = delta U - 2RT, we get delta H < delta U.

Question 30

An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the initial temperature and Tf is the final temperature, which of the following statements is correct?

Accepted Answer

(Tf)irrev > (Tf)rev. In an irreversible expansion, the gas does work on the surroundings without the ability to return to equilibrium, resulting in a higher final temperature compared to a reversible process where the system remains in equilibrium throughout. This leads to the conclusion that the final temperature after irreversible expansion is greater than that after reversible expansion.

Question 31

The standard enthalpy of formation (ΔfH°) at 298 K for methane, CH4(g) is −74.8 kJ mol⁻¹. The additional information required to determine the average energy for C − H bond formation would be

Accepted Answer

the dissociation energy of H2 and enthalpy of sublimation of carbon. To calculate the average energy for C−H bond formation, we need the dissociation energy of H2 to understand the energy required to form hydrogen atoms, and the enthalpy of sublimation of carbon to convert solid carbon into gaseous carbon atoms, which are both essential for forming methane.

Question 32

The enthalpy changes for the following processes are listed below:
Cl2(g) = 2Cl(g), 242.3 kJ mol⁻¹
I2(g) = 2I(g), 151.0 kJ mol⁻¹
ICl(g) = I(g) + Cl(g), 211.3 kJ mol⁻¹
I2(s) = I2(g), 62.76 kJ mol⁻¹
Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard ent

Accepted Answer

+16.8 kJ mol⁻¹. Form ICl from elements: 1/2 Cl2(g)->Cl = 121.15, 1/2 I2(g)->I = 75.5, 1/2 I2(s)->1/2 I2(g) = 31.38, and I+Cl->ICl = -211.3. Sum = +16.8 kJ/mol.

Question 33

(ΔH − ΔU) for the formation of carbon monoxide (CO) from its elements at 298 K is (R = 8.314 J K⁻¹ mol⁻¹)

Accepted Answer

1238.78 J mol⁻¹. For C(s) + 1/2 O2(g) -> CO(g), delta n_g = 1 - 1/2 = +1/2. So delta H - delta U = delta n_g * R * T = 0.5 * 8.314 * 298 = +1238.78 J/mol.

Question 34

In conversion of lime-stone to lime, CaCO3(s) → CaO(s) + CO2(g), the values of ΔH° and ΔS° are +179.1 kJ mol⁻¹ and 160.2 J/K mol⁻¹ respectively at 298 K and 1 bar. Assuming that ΔH° and ΔS° do not change with temperature, the temperature above which conversion of limestone to lime will be

Accepted Answer

1118 K. The conversion of limestone to lime becomes spontaneous when the Gibbs free energy change (ΔG) is negative, which occurs when the temperature exceeds the threshold calculated using the equation ΔG = ΔH - TΔS. By rearranging this equation to find the temperature at which ΔG equals zero, we can determine that the correct temperature is 1118 K.

Question 35

Assuming that water vapour behaves as an ideal gas, the internal energy change (ΔU) when 1 mol of water is vapourised at 1 bar pressure and 100 °C, [given: molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol⁻¹ and R = 8.3 J mol⁻¹ K⁻¹] will be

Accepted Answer

37.904 kJ mol⁻¹. Vaporising 1 mol gives dn_g = +1, so dU = dH - dn_g*RT = 41000 - (1)(8.3)(373) = 37904 J = 37.904 kJ/mol.

Question 36

Which statement about a spontaneous process is correct?

Accepted Answer

In an isolated system, a spontaneous process is accompanied by an increase in entropy.. A spontaneous process in an isolated system tends to increase the overall disorder or entropy, reflecting the natural tendency of systems to evolve towards a state of higher entropy.

Question 37

The standard molar entropies of X2, Y2, and XY3 are 60, 40, and 50 J K−1 mol−1, respectively. For the reaction 1/2 X2 + 3/2 Y2 → XY3 with ΔH = −30 kJ, the temperature at which the system is in equilibrium is

Accepted Answer

750 K. The equilibrium temperature can be determined using the Gibbs free energy equation, which relates enthalpy, entropy, and temperature. In this case, the negative enthalpy change indicates an exothermic reaction, and the calculated change in entropy suggests that the reaction favors the formation of products at 750 K, making it the correct equilibrium temperature.

Question 38

Using the thermochemical information below, and taking ΔfG° for H+(aq) as zero:

H2O(l) → H+(aq) + OH−(aq); ΔH = 57.32 kJ
H2(g) + 1/2 O2(g) → H2O(l); ΔH = −286.20 kJ

What is the standard enthalpy of formation of OH−(aq) at 25 °C?

Accepted Answer

−228.88 kJ. The standard enthalpy of formation of OH−(aq) can be derived from the given reactions by rearranging the equations and applying Hess's law. By combining the enthalpy changes, we find that the formation of OH−(aq) from its elements results in an enthalpy of −228.88 kJ.

Question 39

The standard enthalpy of formation of ammonia, NH3, is −46.0 kJ mol−1. If the enthalpy change for forming H2 from gaseous atoms is −436 kJ mol−1 and that for forming N2 from gaseous atoms is −712 kJ mol−1, what is the average bond enthalpy of an N−H bond in NH3?

Accepted Answer

+352 kJ mol−1. The average bond enthalpy of an N−H bond in NH3 can be calculated by considering the enthalpy changes associated with the formation of NH3 from its elements in their gaseous state. The formation of NH3 releases energy, and by using the given enthalpy changes for the formation of H2 and N2, we can derive the bond enthalpy, which results in the correct answer of +352 kJ mol−1.

Question 40

For a given reversible reaction at temperature T, both ΔH and ΔS are positive. If Tₑ denotes the equilibrium temperature, under what condition will the reaction be spontaneous?

Accepted Answer

T is greater than Tₑ. A reaction is spontaneous when the Gibbs free energy change (ΔG) is negative. Since both ΔH and ΔS are positive, the reaction will be spontaneous at temperatures higher than the equilibrium temperature (Tₑ), where the entropy increase can outweigh the enthalpy increase, leading to a negative ΔG.

Question 41

For the reaction

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

at 27 °C, the enthalpy change is −1366.5 kJ mol−1. What is the corresponding internal energy change at this temperature?

Accepted Answer

−1364.0 kJ. delta n_gas = 2 - 3 = -1. dU = dH - (delta n)RT = -1366.5 - (-1)(8.314e-3 kJ)(300) = -1366.5 + 2.49 = -1364.0 kJ. Correct option: -1364.0 kJ.

Question 42

A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0°C. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be: (R = 8.314 J/mol K) [ln 7.5 = 2.01]

Accepted Answer

q = +208 J, w = −208 J. Heat absorbed q = +208 J. For an isothermal process of an ideal gas dU = 0, so w = -q = -208 J. Thus q = +208 J, w = -208 J.

Question 43

For complete combustion of ethanol, C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l), the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol−1 at 25°C. Assuming ideality the enthalpy of combustion, ΔcH, for the reaction will be: (R = 8.314 kJ mol−1)

Accepted Answer

−1366.95 kJ mol−1. Bomb calorimeter gives dU = -1364.47 kJ/mol at constant volume. dn_g = 2 - 3 = -1, so dH = dU + dn_g*R*T = -1364.47 + (-1)(0.008314)(298) = -1366.95 kJ/mol.

Question 44

Given that the heats of combustion of carbon and carbon monoxide are −393.5 kJ mol−1 and −283.5 kJ mol−1, respectively, what is the heat of formation (in kJ) of one mole of carbon monoxide?

Accepted Answer

−110.5. C + O2 -> CO2 is -393.5; CO + 1/2 O2 -> CO2 is -283.5. For C + 1/2 O2 -> CO, dHf = -393.5 - (-283.5) = -110.5 kJ/mol.

Question 45

ΔU is equal to

Accepted Answer

Adiabatic work. In an adiabatic process q = 0, so by the first law delta-U = q + w = w; thus delta-U equals the adiabatic work. (In an isothermal process of an ideal gas delta-U = 0.) Correct option: Adiabatic work.

Question 46

Given
C(graphite) + O2(g) → CO2(g); ΔrH° = −393.5 kJ mol−1
H2(g) + 1/2 O2(g) → H2O(l); ΔrH° = −285.8 kJ mol−1
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g); ΔrH° = +890.3 kJ mol−1
Based on the above thermochemical equations, the value of ΔfH° at 298 K for the reaction C(graphite) + 2H2(g) → CH4(g) wi

Accepted Answer

−74.8 kJ mol−1. C + 2H2 -> CH4 = (C+O2->CO2) + 2(H2+1/2O2->H2O) + (CO2+2H2O->CH4+2O2) = (-393.5) + 2(-285.8) + (+890.3) = -74.8 kJ/mol. So dfH = -74.8 kJ/mol.

Question 47

The combustion of benzene (l) gives CO2 (g) and H2O (l). Given that heat of combustion of benzene at constant volume is −3263.9 kJ mol−1 at 25°C; heat of combustion of benzene at constant pressure will be: (R = 8.314 J K−1 mol−1)

Accepted Answer

−3267.6. For C6H6(l) + 7.5 O2(g) -> 6 CO2(g) + 3 H2O(l), dng = 6 - 7.5 = -1.5. dH = dU + dng*RT = -3263.9 + (-1.5)(8.314)(298)/1000 = -3263.9 - 3.72 = -3267.6 kJ/mol.

Question 48

For the process A(l) → 2B(g), the internal energy change is ΔU = 2.1 kcal and the entropy change is ΔS = 20 cal K⁻¹ at 300 K. The Gibbs free energy change, in kcal, is:

Accepted Answer

−2.70. dH = dU + (delta n)RT = 2.1 + 2(1.987e-3)(300) = 2.1 + 1.19 = 3.29 kcal. dG = dH - T dS = 3.29 - 300(0.020) = 3.29 - 6.0 = -2.70 kcal.

Question 49

For a reaction that occurs on its own, the values of ΔG, the equilibrium constant (K), and E°cell are, respectively,

Accepted Answer

negative, greater than 1, positive. A spontaneous reaction is characterized by a negative ΔG, indicating that it can occur without external energy. A greater than 1 equilibrium constant (K) suggests that the products are favored at equilibrium, while a positive E°cell reflects a favorable electrochemical reaction, supporting the spontaneity of the process.

Question 50

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l)
At 298 K standard Gibbs energies of formation for CH3OH(l), H2O(l) and CO2(g) are -166.2, -237.2 and -394.4 kJ mol⁻¹ respectively. If standard enthalpy of

Accepted Answer

97%. The efficiency of the fuel cell is calculated by comparing the Gibbs free energy change of the reaction to the enthalpy change. Given the high Gibbs energy change relative to the enthalpy of combustion, the fuel cell operates at a high efficiency, which is reflected in the calculated value of 97%.

JEE Main Chemistry: Thermodynamics questions with solutions

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