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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

Write 1 Faraday in terms of coulombs.(a) 96500 C(b) 95600 C(c) 9560 C(d) 9650 CThis question was addressed to me in an interview.I'd like to ask this question from Gibbs Energy Change and Equilibrium in section Thermodynamics of Chemistry – Class 11

Answer»

Right answer is (a) 96500 C

The best I can explain: FARADAY is a unit which has no dimensions and it represents electric charge quantity. In the International System of units, it’s coulombs. 1 Faraday is equal to 96500 coulombs. F represents Faraday constant.

2.

In the equation, ΔG = – 2.303 RT logK, what is K?(a) change in temperature(b) kelvin(c) equilibrium constant(d) change in enthalpyThe question was asked during a job interview.I'm obligated to ask this question of Gibbs Energy Change and Equilibrium in division Thermodynamics of Chemistry – Class 11

Answer»

Correct CHOICE is (c) equilibrium constant

The best I can explain: In the above equation ΔG = – 2.303 RT logK, G is the change in standard Gibbs free energy, R is the universal gas constant, T is TEMPERATURE, K is the equilibrium constant while LOG stands for the logarithm of base 10.

3.

What is the relation between Gibbs free energy and the EMF of the cell?(a) ΔG = -nFEcell(b) G = -nFEcell(c) ΔG = -nEcell(d) ΔG = -nFcellI had been asked this question in an international level competition.Query is from Gibbs Energy Change and Equilibrium in chapter Thermodynamics of Chemistry – Class 11

Answer»

Correct ANSWER is (a) ΔG = -nFEcell

Easy explanation: The RELATION between Gibbs free ENERGY and EMF of the cell is ΔG = -nFEcell, where G is the caps free energy and as a number of electrons lost or gained while f depicts Faraday and E cell STAND for standard electrode POTENTIAL.

4.

Calculate the Gibbs free energy for the conversion of oxygen to Ozone at room temperature if KP is given as 2.47 x 10^-29.(a) 163 kJ/mol(b) 163 J/mol(c) 163 kJ(d) 163 k/molI have been asked this question in class test.My question comes from Gibbs Energy Change and Equilibrium topic in chapter Thermodynamics of Chemistry – Class 11

Answer»

The correct choice is (a) 163 kJ/mol

Explanation: The chemical EQUATION for the conversion of oxygen to Ozone is 3/2O2 → O3. We have the equation, ΔG = – 2.303 RT LOG KP. So by SUBSTITUTING we get ΔG = – 2.303 x 8.314 J/K-mol x 293K x 2.47 x 10^-29 = 163000 J/mol = 163 kJ/mol.

5.

The Gibbs free energy is positive when a change in enthalpy and change in entropy are positive at ____________(a) high temperatures(b) low temperature(c) all temperatures(d) only at 0 KelvinThis question was posed to me in homework.This interesting question is from Gibbs Energy Change and Equilibrium topic in portion Thermodynamics of Chemistry – Class 11

Answer»

Right choice is (b) low temperature

Explanation: When a CHANGE in entropy and the change in enthalpy are positive the GIBBS FREE energy is positive at low TEMPERATURES and negative at HIGH temperatures. We can obtain this through the equation ΔG = ΔH – TΔS.

6.

Which of the following is not a type of exergonic reaction?(a) formation of table salt from Sodium and chlorine(b) combustion reaction(c) chemiluminescence(d) photosynthesisI have been asked this question in class test.I'm obligated to ask this question of Gibbs Energy Change and Equilibrium in chapter Thermodynamics of Chemistry – Class 11

Answer»

Right choice is (d) photosynthesis

To explain: A reaction that is spontaneous and favorable is known as an exergonic reaction, it RELEASES energy into the surroundings. The system’s FREE energy decreases here. The GIBBS free energy of an exergonic reaction is NEGATIVE.

7.

The melting of ice into liquid water is an example of tube _______________ reaction.(a) endergonic(b) exergonic(c) exothermic(d) endothermicThis question was addressed to me in an internship interview.The above asked question is from Gibbs Energy Change and Equilibrium topic in division Thermodynamics of Chemistry – Class 11

Answer»

The correct answer is (a) endergonic

To explain: An unfavorable REACTION or a nonspontaneous reaction that requires more energy than we get from that reaction is called endergonic reaction. It OBSERVES heat from SURROUNDINGS and the change in ENTROPY decreases it is ALSO a type of endothermic reaction.

8.

In a reaction, the change in entropy is given as 2.4 cal/K and the change in Gibbs free energy is given as 3.4 kcal, calculate the change in heat at the temperature of 20-degree centigrade?(a) 3.4 kcal(b) 3.4 cal(c) 3.4 kJ(d) 3.4 JI got this question by my college professor while I was bunking the class.My question is taken from Gibbs Energy Change and Equilibrium topic in chapter Thermodynamics of Chemistry – Class 11

Answer»

Correct OPTION is (a) 3.4 kcal

Explanation: Using the EQUATION ΔG = ΔH – TΔS, where G is represented by the Gibbs free energy, H is represented by the CHANGE in enthalpy, T is represented by the temperature and S is represented by the entropy. ΔH = ΔG + TΔS = 3.4 kcal – 293 x 2.4 cal/K = 3.4 kcal.

9.

Calculate the Gibbs free energy for the reaction of conversion of ATP into ADP at 293 Kelvin the change in enthalpy is 19.07 Kcal and the change in entropy is 90 cal per Kelvin.(a) 7.3 cal(b) -5.3 Kcal(c) 7.3 Kcal(d) -7.3 KcalI got this question during an internship interview.I'm obligated to ask this question of Gibbs Energy Change and Equilibrium in section Thermodynamics of Chemistry – Class 11

Answer»

Right ANSWER is (d) -7.3 Kcal

To EXPLAIN: We have ΔG = ΔH – TΔS; by SUBSTITUTING ΔH = 19.07 kcal and ΔS = 90 cal/K, we get ΔG = 19.07 Kcal – 293(90 cal/K) = 19.07 Kcal – 26.37 Kcal = -7300 cal = -7.3 Kcal. The Gibbs FREE energy CHANGE is -7.3 Kcal.

10.

Find out whether the following reaction is spontaneous or not at 127 degrees centigrade? N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g); ΔH = 92.22 kJ/mol and ΔS = -198.75 J/K-mol.(a) it is spontaneous(b) it is not spontaneous(c) it may be spontaneous(d) cannot predictThis question was posed to me by my school teacher while I was bunking the class.The doubt is from Gibbs Energy Change and Equilibrium in section Thermodynamics of Chemistry – Class 11

Answer» RIGHT ANSWER is (b) it is not spontaneous

Explanation: We have ΔG = ΔH – TΔS; by substituting ΔH = 92.22 kJ/mol and ΔS = -198.75 J/K-mol, we get ΔG = 92.22 kJ/mol – 400k(-198.75 J/K-mol) = 92.22 kJ/mol + 79.5 kJ/mole = 171.72 KJ/mol. As Gibbs free ENERGY is positive the reaction is NON spontaneous.
11.

ΔS(universe) > 0 can also be written as _________(a) ΔS(system) + ΔS(surroundings) > 0(b) ΔS(surroundings) > 0(c) ΔS(system)(d) ΔS(system) – ΔS(surroundings) > 0I had been asked this question in an international level competition.My question is from Thermodynamics in section Thermodynamics of Chemistry – Class 11

Answer»

The correct option is (a) ΔS(system) + ΔS(surroundings) > 0

For EXPLANATION I WOULD SAY: ΔS(universe) > 0 can also be written as ΔS(system) + ΔS(surroundings) > 0, as the universe is nothing but the system and surroundings. The above mentioned CONDITION is used when a reaction is at equilibrium.

12.

The entropy of the universe is always increasing is ____________(a) zeroth law of thermodynamics(b) first law of thermodynamics(c) second law of thermodynamics(d) third law of thermodynamicsI have been asked this question by my college professor while I was bunking the class.My query is from Thermodynamics in portion Thermodynamics of Chemistry – Class 11

Answer»

Correct answer is (B) first LAW of thermodynamics

For explanation I would say: The second law of thermodynamics is given as the entropy of the universe is always increasing in the course of every spontaneous or NATURAL CHANGE also can be SAID as the heat cannot flow itself from a colder to a hotter body.

13.

Third law of thermodynamics is only applicable for perfectly crystalline substances.(a) true(b) falseThe question was posed to me during a job interview.My question comes from Thermodynamics in section Thermodynamics of Chemistry – Class 11

Answer»

Right option is (a) true

The explanation: We can define the THIRD law of thermodynamics ENTROPY of a perfectly crystalline substance at zero kelvin or absolute zero is taken to be zero, but this is not APPLICABLE if there is any IMPERFECTION at 0 k, the entropy will be larger than 0.

14.

Reaction is spontaneous if Gibbs free energy is __________(a) greater than zero(b) equal to zero(c) less than zero(d) infinityI had been asked this question in an interview for job.This key question is from Thermodynamics in portion Thermodynamics of Chemistry – Class 11

Answer»

The CORRECT option is (c) LESS than zero

Explanation: A reaction is spontaneous if Gibbs free ENERGY is less than zero and the reaction is nonspontaneous if the Gibbs free energy is GREATER than zero. The reaction is at equilibrium STATE if gives free energy is equal to zero.

15.

If the enthalpy is positive and entropy is negative, what do you understand about Gibbs free energy?(a) it is positive(b) negative(c) may be negative(d) may be positiveThis question was addressed to me during an online exam.I'm obligated to ask this question of Thermodynamics topic in chapter Thermodynamics of Chemistry – Class 11

Answer»

The CORRECT choice is (a) it is positive

To elaborate: According to the Helmholtz equation; G = H -TS, when enthalpy H is positive and entropy S is negative, the GIBBS free energy is always positive and the REACTION is not spontaneous at all temperatures.

16.

Which of the following relation is true between Gibbs energy, enthalpy, temperature, and entropy?(a) G = H -TS(b) G = H -T(c) G = H -S(d) G = -TSI have been asked this question during a job interview.My question comes from Thermodynamics topic in division Thermodynamics of Chemistry – Class 11

Answer» RIGHT option is (a) G = H -TS

The best explanation: The energy that is available for a system at some CONDITIONS and by which useful work can be done is Gibbs FREE energy. The relation between Gibbs free energy, enthalpy, temperature and entropy is GIVEN by Gibbs Helmholtz equation; G = H -TS.
17.

Entropy is an ___________ property.(a) intensive(b) extensive(c) neither intensive or extensive(d) both intensive and extensiveI got this question during an interview.My query is from Thermodynamics topic in chapter Thermodynamics of Chemistry – Class 11

Answer»

Correct option is (B) extensive

Explanation: Entropy is a measurement of RANDOMNESS or disorder of MOLECULES it is a state function as WELL as an extensive property. Its units are J/K-mole. Extensive property depends on the SIZE and quantity of the mass.

18.

Spontaneous reactions that occur or mostly ________in nature.(a) endothermic(b) exothermic(c) both endothermic and exothermic(d) neither exothermic nor endothermicThe question was posed to me during an interview.The above asked question is from Thermodynamics topic in chapter Thermodynamics of Chemistry – Class 11

Answer» CORRECT CHOICE is (b) exothermic

Explanation: All the PROCESSES which occur with a decrease of energy are exothermic in nature, those with change of enthalpy as negative occurs spontaneously. It is not true in case of some endothermic REACTIONS, when they occur spontaneously.
19.

Entropy increases for a spontaneous reaction.(a) true(b) falseI got this question during an online interview.This intriguing question originated from Thermodynamics topic in portion Thermodynamics of Chemistry – Class 11

Answer» RIGHT answer is (a) TRUE

For explanation: A process is spontaneous if and only if the entropy of Universe increases for a process to be spontaneous ΔS(universe) > 0. At equilibrium, ΔS = 0. Saudi about the statement that the Entropy increase for a spontaneous REACTION is true.
20.

Which of the following is not a spontaneous process?(a) sugar dissolves in water(b) melting of iron(c) rusting of iron(d) evaporation of waterThe question was asked in an online interview.Query is from Thermodynamics topic in chapter Thermodynamics of Chemistry – Class 11

Answer»

The correct answer is (b) melting of IRON

To explain I would say: A physical or chemical process which occurs by its on in a particular direction under a particular set of conditions without any force is KNOWN as a spontaneous process and it cannot be reversed. Here melting of iron is only POSSIBLE when we SUPPLY a large amount of heat so it is not a spontaneous process.

21.

The enthalpy of dilution of a solution is __________ on the original concentration of the solution and the amount of solvent added.(a) dependent(b) independent(c) may be dependent(d) may be independentThis question was posed to me during an online interview.This interesting question is from Thermodynamics topic in portion Thermodynamics of Chemistry – Class 11

Answer»

The correct choice is (a) dependent

Easiest explanation: Enthalpy of dilution is the enthalpy change when 1 mole of a SUBSTANCE is diluted from ONE concentration to another. So it is dependent on your original concentration of the solution and the AMOUNT of solvent ADDED.

22.

Enthalpy of solution can either be positive or ________(a) negative(b) does not exist(c) 0(d) infinityThe question was posed to me in an interview for job.I would like to ask this question from Thermodynamics topic in chapter Thermodynamics of Chemistry – Class 11

Answer»

Right answer is (a) negative

Easy explanation: Enthalpy change when 1 mole of a substance is dissolved in a large excess of solvent so that on further delusion no APPRECIABLE heat change occurs, is KNOWN as enthalpy of solution. It can EITHER be positive or negative DEPENDING on the reaction, whether it is exothermic or ENDOTHERMIC.

23.

Which of the following cycle allows has to analyze reaction energies?(a) carbon cycle(b) born Haber cycle(c) nitrogen cycle(d) chemical cycleThe question was posed to me in class test.Question is from Thermodynamics in chapter Thermodynamics of Chemistry – Class 11

Answer»

The CORRECT ANSWER is (b) BORN Haber cycle

To EXPLAIN: An indirect method, in which we can construct an enthalpy diagram is CALLED a Born-Haber cycle as it is nearly impossible to determine lattice enthalpies directly by experiment. It is an approach to analyze reaction energies.

24.

In which of the following reactions the energy released is not an enthalpy of atomization?(a) decomposition of a hydrogen molecule into two hydrogen atoms(b) decomposition of an oxygen molecule into 2 oxygen atoms(c) formation of a water molecule(d) decomposition of chlorine moleculeThe question was posed to me during an online exam.This key question is from Thermodynamics topic in chapter Thermodynamics of Chemistry – Class 11

Answer»

Right answer is (c) formation of a water molecule

To explain: The ENTHALPY of atomization is an enthalpy change that occurs when one molecule of a molecule BREAKS into its atoms, but here the of a water molecule formation does not SATISFY the definition of the enthalpy of atomization.

25.

Decomposition of sodium chloride into a sodium ion and chloride ion releases energy of 788 KJ per Mol, what is the energy called?(a) lattice energy(b) translation energy(c) dilution energy(d) neutralization energyI got this question during an interview for a job.My query is from Thermodynamics topic in portion Thermodynamics of Chemistry – Class 11

Answer»

Correct CHOICE is (a) lattice energy

Easiest explanation: Lattice enthalpy is the enthalpy CHANGE when one molecule of an ionic COMPOUND DISSOCIATES into its IONS and gaseous state. Here sodium chloride is an ionic compound and it decomposes into sodium and chloride ions, so the energy released is lattice energy.

26.

Calculate the carbon carbon double bond energy in ethane from the following reaction, H2C=CH2(g) + H2(g) –> H3C−CH3(g) ΔH = −138 kJ/mol. If Bond enthalpies are: C−C = 348; H−H = 436; C−H = 412 in KJ/mol.(a) 498 KJ/mol(b) 593 KJ/mol(c) 508 KJ/mol(d) 598 KJ/molThis question was posed to me by my college director while I was bunking the class.This interesting question is from Thermodynamics in division Thermodynamics of Chemistry – Class 11

Answer»

Right option is (d) 598 KJ/MOL

Easiest explanation: By following the rule ΔHr = Σ ΔHf[products] – ∑ ΔHf[reactants], we get -(x + 4(412) + 432) + (348 + 6(412)) = -138 KJ/mol; x = 598 KJ/mol. So the CARBON carbon DOUBLE bond energy in Ethane is given as 598 KJ per Mol.

27.

In case of the decomposition of hydrogen molecule into two hydrogen atoms, the enthalpy of atomization is same as the _________(a) bond dissociation enthalpy(b) enthalpy of formation(c) enthalpy of combustion(d) enthalpy of sublimationThe question was posed to me during an interview for a job.My enquiry is from Thermodynamics in chapter Thermodynamics of Chemistry – Class 11

Answer»

Right OPTION is (a) BOND DISSOCIATION enthalpy

For explanation: In case of the decomposition of hydrogen molecule the enthalpy is the same as enthalpy of atomization as well as bond dissociation enthalpy because the bond association enthalpy refers to the breakage of H-H Bond. So that TWO hydrogen atoms are formed whereas enthalpy of atomization is the breakage of hydrogen molecules in ORDER to form two atoms.

28.

The enthalpy of 435 KJ per Mol is obtained from the conversion of hydrogen molecule into two hydrogen atoms.(a) True(b) FalseThis question was posed to me in examination.This question is from Thermodynamics topic in portion Thermodynamics of Chemistry – Class 11

Answer»

Right choice is (a) True

To explain I would SAY: The enthalpy CHANGE that occurs when one mole of a molecule breaks into its atoms is KNOWN as enthalpy of atomization. The energy released during the CONVERSION of a HYDROGEN molecule into two hydrogen atoms is enthalpy of atomization.

29.

ΔHr = Σ ΔHf[products] – Σ ΔHf[reactants].(a) true(b) falseThe question was asked in an interview.My question is based upon Thermodynamics topic in section Thermodynamics of Chemistry – Class 11

Answer»

Right OPTION is (a) TRUE

Best explanation: The EQUATION ΔHr = Σ ΔHf[products] – Σ ΔHf[reactants] says that the enthalpy of a reaction is the DIFFERENCE between the enthalpy of products and enthalpy of reactants. The above statement regarding enthalpy is true.

30.

Calculate the heat of combustion of ethane, in the reaction C2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l) where the heats of formation of ethane gas, carbon dioxide gas and water liquid are –84.7 kJ mol^-1, -393.5 kJ mol^-1 and –285.8 kJ mol^-1 respectively.(a) -1559.7 kJ mol^-1(b) -155.7 kJ mol^-1(c) -159.7 kJ mol^-1(d) -559.7 kJ mol^-1This question was addressed to me by my college director while I was bunking the class.Asked question is from Thermodynamics topic in chapter Thermodynamics of Chemistry – Class 11

Answer» CORRECT answer is (a) -1559.7 kJ mol^-1

To explain: GIVEN that C2H6(g) + 3 1/2 O2(g) → 2CO2(g) + 3H2O(l) As we know that, ΔHr = Σ ΔHf[products] – Σ ΔHf[REACTANTS]; ΔHc(C2H6) = 2 X (-393.5 kJ mol^-1) + 3 X (-285.8 kJ mol^-1) – (- 84.7 kJ mol^-1) – 2 X (0 kJ mol^-1) = ΔHc(C2H6) = – 1559.7 kJ mol^-1.
31.

Enthalpy of combustion as always ________(a) positive(b) negative(c) 0(d) infinityI had been asked this question in my homework.I would like to ask this question from Thermodynamics topic in portion Thermodynamics of Chemistry – Class 11

Answer»

Right OPTION is (B) negative

The EXPLANATION: Enthalpy of combustion is the enthalpy change that takes PLACE when one mole of a compound undergoes complete combustion in the presence of oxygen. It is represented by ΔHc, it is always negative because the PROCESS of combustion is exothermic.

32.

Which of the following is not an application of Hess’s law?(a) determination of heat of formation(b) determination of heat of transition(c) determination of Gibb’s energy(d) determination of heat of hydrationThis question was posed to me during an interview for a job.My doubt stems from Thermodynamics in division Thermodynamics of Chemistry – Class 11

Answer»

The correct answer is (c) determination of Gibb’s energy

Explanation: The following are the applications of Hess’s LAW; determination of heat of FORMATION, determination of heat of transition and determination of heat of HYDRATION, ALSO to CALCULATE bond energies.

33.

What is the unit of standard enthalpy of fusion or molar enthalpy of fusion?(a) KJ Mol(b) KJ per Mol(c) Mol per KJ(d) 1/KJ MolThe question was asked in a job interview.My question is taken from Thermodynamics in division Thermodynamics of Chemistry – Class 11

Answer»

The correct choice is (b) KJ per Mol

To explain: The enthalpy change that OCCURS during melting of one mole of a solid SUBSTANCE in the standard state is called standard enthalpy of fusion or MOLAR enthalpy of fusion, it is represented by the symbol ΔfusH^–, the UNITS of this are KJ per Mol.

34.

Consider the equation 2 H2 + O2 → 2 H2O, what does the 2 in the coefficient of H2O molecule represent?(a) number of particles(b) the number of molecules(c) number of moles(d) number of atomsThis question was addressed to me in a job interview.Query is from Thermodynamics in division Thermodynamics of Chemistry – Class 11

Answer»

The CORRECT choice is (c) number of moles

The EXPLANATION is: In a balanced thermochemical equation, the coefficients always REFER to the number of the moles (but never molecules) of reactants and products involved in a REACTION so 2 in the COEFFICIENT of H2O refers to the number of the moles of water.

35.

When a chemical reaction is reversed the value of enthalpy is reversed in sign.(a) true(b) falseThis question was posed to me in a job interview.I would like to ask this question from Thermodynamics topic in division Thermodynamics of Chemistry – Class 11

Answer»

Right option is (a) TRUE

To explain I WOULD say: For example, the formation of AMMONIA has an enthalpy of -91.8 KJ per MOLE and the decomposition of ammonia has an enthalpy of + 91.8 KJ per Mol. So the above statement that when a chemical reaction is reversed the value of enthalpy is reversed in the sign is true.

36.

What is a change in energy if 18 grams of water is heated from room temperature to 20 degrees above it?(a) 1.50 KJ(b) 0.506 KJ(c) 1.06 KJ(d) 1.506 KJThis question was addressed to me during an internship interview.This intriguing question originated from Thermodynamics in division Thermodynamics of Chemistry – Class 11

Answer»

Correct ANSWER is (d) 1.506 KJ

Easy explanation: We know that Q = msΔT, where Q is the energy,m is the mass of WATER, s is the specific HEAT of water and T is the temperature. So the change in energy REQUIRED here = 18 g x 4.184 J/g-K x 20K = 1.506KJ.

37.

Calculate the internal energy change when 2 moles of water at 0 degrees converts into ice at 0-degree centigrade?(a) 12 KJ per mole(b) 6 KJ per mole(c) 1 KJ per mole(d) 102 KJ per moleThis question was posed to me in a national level competition.My question comes from Thermodynamics in section Thermodynamics of Chemistry – Class 11

Answer» CORRECT option is (a) 12 KJ PER mole

Easiest explanation: ENERGY change when 1 MOL of water at 0-degree centigrade changes into ice at 0 degrees in centigrade is 6 kJ/mol, So the internal energy change when 2 moles of water at 0 degrees CONVERTS into ice at 0 degrees is 12 kJ/mole.
38.

Consider that, a ball is immersed in water at room temperature and then taken out having 18 grams of water on it, how much amount of energy is required to dry that water at room temperature?(a) 41.43 KJ/mol(b) 49.53 KJ/mol(c) 41.3 KJ/mol(d) 41.53 KJ/molThe question was asked by my college professor while I was bunking the class.I need to ask this question from Thermodynamics topic in chapter Thermodynamics of Chemistry – Class 11

Answer»

Correct choice is (d) 41.53 KJ/mol

Best EXPLANATION: HEAT required to eliminate water : n x ΔvapH^– = (1 mol) × (44.01 kJ mol^–1) = 44.01 kJ mol^-1. Δvapor = ΔvapH – ΔnRT = 44.01 kJ mol^-1 – 1×8.314 J/K-mol x 298 k x 10^-3 = 41.53 KJ/mol. So the amount of energy NEEDED is 41.53 KJ/mol.

39.

The standard state of a substance is considered when the temperature is 298 k and the pressure is ____________(a) 1 ATM(b) 1 bar(c) 1 Pascal(d) 760 mm HGThis question was posed to me during an interview.My question comes from Thermodynamics topic in division Thermodynamics of Chemistry – Class 11

Answer» CORRECT CHOICE is (b) 1 BAR

Easy explanation: All the standard STATES of a substance are considered when the temperature is 298 Kelvin and the pressure is 1 bar. 1 bar = 0.987 atmospheric pressure = 10000 Pascal = 750.0617 mm of Mercury.
40.

All the enthalpies of fusion are positive.(a) true(b) falseThe question was posed to me in exam.Question is taken from Thermodynamics topic in chapter Thermodynamics of Chemistry – Class 11

Answer» RIGHT answer is (a) true

The best explanation: Fusion is a PROCESS of conversion of liquid to solid the enthalpy is energy that is required for a process. As the melting of a solid is endothermic, the enthalpies of fusion are POSITIVE so the above statement is true.
41.

The enthalpy, internal energy during a process and change in volume are 500 units, 400 units, and 2 units. What is the pressure that is exerted on the gas during this process?(a) 20 units(b) 80 units(c) 100 units(d) 50 unitsThe question was posed to me in an interview.This is a very interesting question from Thermodynamics topic in division Thermodynamics of Chemistry – Class 11

Answer»

The correct answer is (d) 50 UNITS

Explanation: We know thatΔH = ΔU + PΔV; ΔH is the enthalpy, ΔU is the internal energy, ΔV is the CHANGE in volume and P is the PRESSURE. So by SUBSTITUTING the enthalpy, internal energy during a process and change in volume as 500 units, 400 units and 2 units, we get pressure as 50 units.

42.

The total heat energy utilized for increasing the temperature by 4 degrees Kelvin in a 3 kgs substance is 100 KJ what is the specific heat capacity of that substance?(a) 8.34 KJ/g-k(b) 8.34 KJ/Kg-k(c) 8.34 KJKg-k(d) 8.34 KJ/KgThe question was posed to me in examination.The question is from Thermodynamics in division Thermodynamics of Chemistry – Class 11

Answer»

Right ANSWER is (b) 8.34 KJ/Kg-k

The best I can explain: The formula of heat ENERGY is GIVEN by the EXPRESSION: Q = mcΔT, m is the mass of the substance, c is the specific heat CAPACITY and ΔT is the temperature difference. c = Q/mΔT; c = 100KJ/3kg(4k) = 8.34 KJ/Kg-k.

43.

When 1 kg of water at 373 k, is converted into steam how much amount of heat energy is required?(a) 22600 KJ(b) 226 KJ(c) 2260 KJ(d) 22.6 KJI have been asked this question in an interview for internship.This interesting question is from Thermodynamics topic in portion Thermodynamics of Chemistry – Class 11

Answer»

Right option is (c) 2260 KJ

For EXPLANATION: The latent HEAT of VAPORIZATION of water is 2260 KJ/Kg. The heat that is REQUIRED to convert water at 373 K to steam is given by Q = mL, where m = mass of the water and L = latent heat vaporization of water; heat energy required = 1 kg x 2260 KJ/Kg = 2260 KJ.

44.

What is the value of specific heat capacity in the adiabatic process?(a) 0(b) infinity(c) positive(d) negativeI had been asked this question in a job interview.Origin of the question is Thermodynamics topic in portion Thermodynamics of Chemistry – Class 11

Answer»

Correct answer is (a) 0

To ELABORATE: During an adiabatic PROCESS the change in TOTAL energy is ZERO, As we know that the specific heat capacity is given by the total heat required by mass X change in temperature, heat energy is zero the specific heat capacity becomes zero.

45.

During the process of conversion of ice into the water, the specific heat capacity is given by _______(a) 0(b) positive(c) infinity(d) negativeThis question was addressed to me by my school principal while I was bunking the class.My question is taken from Thermodynamics in portion Thermodynamics of Chemistry – Class 11

Answer»

The correct answer is (c) infinity

For explanation I would SAY: During the phase CHANGE of a substance the temperature change is zero. As we KNOW that specific heat CAPACITY = Q/cΔT which is zero; specific heat capacity becomes infinity, so during the process of conversion of ice into the WATER, the specific heat capacity is infinity.

46.

“c” the specific heat capacity of a substance is given by temperature difference is given by ΔT and the heat energy is given by Q then what is the mass of the substance?(a) Q/cΔT(b) cΔT /Q(c) QcΔT(d) QΔTThis question was addressed to me in class test.Origin of the question is Thermodynamics topic in section Thermodynamics of Chemistry – Class 11

Answer» RIGHT option is (a) Q/cΔT

For explanation I would say: Heat ENERGY of a substance is denoted by Q and is given by the expression mcΔT, where m is a mass of the substance, c is a specific heat capacity and ΔT is TEMPERATURE difference, so the mass of the substance is Q/cΔT.
47.

Bomb calorimeter is used to determine ____________(a) molar heat capacity(b) heat of combustion(c) rate kinetics(d) affinityI got this question during an interview.The question is from Thermodynamics topic in section Thermodynamics of Chemistry – Class 11

Answer»

Correct ANSWER is (B) heat of combustion

Easiest explanation: A bomb calorimeter is used to measure the heat of combustion of a reaction. It has to withstand a large amount of PRESSURE, in ORDER to determine the heat of combustion. It is an isochoric process and the heat energy is EQUAL to the internal energy.

48.

Heat capacity of a Bomb calorimeter is given by _______(a) CV(b) CP(c) CM(d) CBThe question was asked in homework.Enquiry is from Thermodynamics in section Thermodynamics of Chemistry – Class 11

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The correct CHOICE is (a) CV

To explain I WOULD say: A BOMB calorimeter operates at a constant volume i.e. it is an isochoric process. So the heat capacity is CV which represents heat capacity at constant volume. While CP represents heat capacity at constant PRESSURE.

49.

If an exothermic reaction occurs in a Bomb calorimeter then the temperature of the water bath ________(a) increases(b) decreases(c) remains constant(d) cannot predictThis question was posed to me in an online interview.My question is based upon Thermodynamics topic in section Thermodynamics of Chemistry – Class 11

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The correct option is (a) increases

Easiest EXPLANATION: In a Bomb calorimeter, the reaction occurs in a vessel and that is SURROUNDED by the water bath. If it is an exothermic reaction the TEMPERATURE rises and if it is an endothermic reaction the temperature decreases. Temperature can be MEASURED USING a thermometer.

50.

Bomb calorimeter is __________in nature.(a) isothermal(b) isochoric(c) isobaric(d) absoluteThe question was asked during an online exam.The origin of the question is Thermodynamics topic in portion Thermodynamics of Chemistry – Class 11

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Correct CHOICE is (b) isochoric

To elaborate: ABomb calorimeter works at CONSTANT volume, so it is isochoric in nature. Here, the heat energy that’s measured is only the internal energy. The WORK done is zero because of no change in volume.