iit part test 1 thermodynamics by S K Sinha Kota by S.K.sinha See Chemistry Animations at...

SINHA IIT CHEMISTRY JEE-SIMILAR TEST.

1-D10 Talwandi ,B-7. JAWAHAR NAGAR.Main Road ,Kota. Ph-0744-2422383 .Mo-93527-55722

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PAPER-1(Thermodynamics Part test-1) INSTRUCTIONS

A . General 1. This booklet is your Question Paper containing 23 questions. 2. Fill in the boxes provided below on this page and also write your Name and R egistration No. in the space provide on the back page of this booklet. 3. The answer sheet, a machine-readable Objective Response Sheet (ORS), is provided separately. 4. DO NOT TAMPER WITH/MULTILATE THE ORS OR THE BOOKLET. 5. Do not break the seals of the question-paper booklet before being instructed to do so by the invigilators. B . Filling the OR S : 6. On the lower part of the ORS, write in ink, your name in box L1, your Registration No. in box L2 and Name of the Centre in box L3. Do not write these anywhere else. 7. Write your Registration No. in ink, in the box L 4 provided in the lowerpart of the ORS and darken the appropriate bubble UNDER each digit of your Registration No.with a good quality H B pencil. 8. The ORS has a CODE printed on its lower and upper parts. 9. Make sure the CODE on the ORS is the same as that on this booklet and put your signature in ink in box L5 on the ORS affirming that you have verified this. 10. IF THE CODES DO NOT MATCH , ASK FOR A CHANGE OF THE BOOKLET. C . Question paper format : D . Marking scheme 11. For each question in Section I , you will be awarded 3marks if you darkened only the bubble to the correct answ er and zero mark if no bubble is darkened. In all other cases, minus one (–1) mark will be awarded. 12. For each question in Section II , you will be awarded 4 marks if you darkened all the bubble(s) to the correct answer and zero mark for all other cases. It M ay be noted that there is nonegative marking for wrong answer. 13. For each question in Section III , you will be awarded 3 marks if you darken only the bubble to the correct answ er and zeromark if no bubble is darkened.In all other cases, minus one (-1) mark will be awarded . 14. For each question in Section IV , you will be awarded 4 marks if you darken only the bubble to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (-1) mark will be awarded . …………………………………………………………………………………………………………………………

SECTION - I Straight Objective Type

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D),out of which ONLY ONE is correct. ………………………………………………………………………………………………………………………….. 1. When 1 mole of white tin changes to gray tin at 13.0oC, the change in entropy is -7.5 J * K-1. This means that (A) the change from white to gray tin is predicted to be spontaneous on the basis of the entropy change. (B) gray tin has an entropy of -7.5 J * K-1. (C) gray tin is more ordered than white tin. (D) gray tin has a higher molar entropy than white tin. 2. Which of the following has the smallest molar entropy at 298 K? (A) S(s, rhombic) (B) C(s, diamomd) (C) Sn(s, gray) (D) Sn(s, white) 3. For the freezing of water at 0oC, (A) ΔSsurroundings is less than ΔSsystem (B) ΔSsurroundings is greater than ΔSsystem (C) ΔSsurroundings = ΔSsystem (D) (ΔSsurroundings + ΔSsystem) is less than zero

4. If ΔHo = -1202 kJ and ΔSo = -217 J * K-1 for the combustion of 2 moles of magnesium, the change in entropy of the surroundings at 298 K is (A) 217 J * K-1 (B) -217 J K-1 (C) 4.03 ∗ 103 J * K-1 (D) -4.03 ∗ 103 J * K-1 5. Consider the following reaction N2H4(l) + 2H2O2(l) −−> N2(g) + 4H2O(g) If ΔHro = -530 kJ at 298 K, then (A) the reaction will only be spontaneous at very low temperatures. (B) the reaction is spontaneous at all temperatures.



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(C) the reaction is not spontaneous at any temperature. (D) the reaction will only be spontaneous at very high temperatures. 6. Calculate ΔGro for the decomposition of ammonium nitrate at 298 K. NH4NO3(s) −−> N2O(g) + 2H2O(l)

ΔHfo, kJ * mol-1 -365.56 82.05 -285.83

Smo, J * K-1 * mol-1 151.08 219.85 69.91 (A) -186.21 kJ (B) -332.64 kJ (C) +99.62 kJ (D) -124.05 kJ ………………………………………………………………………………………………………

Section - II Multiple Correct Answer Type

This section contains 4 multiple choice answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. ………………………………………………………………………………………………………………………….. 7. The equilibrium constant for the reaction NH4Cl(s) <==> NH3(g) + HCl(g)

is 1.1 × 10-16 at 25oC. If the equilibrium constant is 6.5 ∗ 10-2 at 300oC, which statement is correct? (A) The reaction is exothermic. (B) The reaction is not spontaneous at any temperature. (C) The reaction is spontaneous at all temperatures. (D) The reaction is spontaneous at high temperatures. 8.Estimate the temperature at which mercury(II) oxide can be expected to decompose. 2HgO(s) −−> 2Hg(l) + O2(g)

ΔHfo, kJ * mol-1 -90.83

Smo, J * K-1 * mol-1 70.29 76.02205.14 (A) HgO(s) is unstable and decomposes at all temperatures (B) greater than 839 K (C) HgO(s) does not decompose at any temperature (D)greater than 566oC 9.The purification of nickel by the Mond process uses the reactions below. Ni(s, impure) + 4CO(g) −−> Ni(CO)4(g) (1) Ni(CO)4(g) −−> Ni(s, pure) + 4CO(g) (2) If reaction (1) occurs at 353 K and reaction (2) at 473 K, reactions (1) and (2) are (A) both have ΔG = 0. (B) entropy- and enthalpy-driven, respectively. (C) 1st is enthalpy -driven. (D) enthalpy- and entropy-driven, respectively. 10. Given : Enthalpy of combustion of methane ΔrH = –890.36 kJ mol–1

Enthalpy of combustion of C (graphite) ΔrH = –393.51 kJ mol–1

H2(g) + O2(g) ⎯→ H2O(l) ΔrH = –285.85 kJ mol–1

Enthalpy of dissociation of H2(g) ΔrH = –435.93 kJ mol–1

Enthalpy of sublimation of C (graphite) ΔrH = –716.68 kJ mol–1 The bond enthalpy of C — H from the following data at 298 K is – (A) –415.85 kJ mol–1 (B) 415.85 kJ mol–1 (C) 1663.39 kJmol–1 (D) 166.339 kJ mol–1 ……………………………………………………………………………………………………………….

SECTION - III Reasoning Type

This section contains 4 reasoning questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. …………………………………………………………………………………………………………………………..



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11. STATEMENT-1 :For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero. STATEMENT-2 : At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 © Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 12. . STATEMENT-1 :ΔHhydrogenation of theoretical cyclohexatriene is greater than Benzene. STATEMENT-2 : The difference is called enthalpy of formation. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 © Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 13. . STATEMENT-1 : If rubber band is stretched entropy decreases. STATEMENT-2 : stretching rubber band is non spontaneous. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 © Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 14. . STATEMENT-1 : Internal energy is exact diferential . STATEMENT-2 : Internal energy is sum of two inexact differentials (ibternal energy and work.) (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 © Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True ……………………………………………………………………………………………………………………..

SECTION – IV Linked Comprehension Type

This section contains 3 paragraphs C15-17, C18-20 and C21-23. Based upon each paragraph, 3 multiple choice question have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. …………………………………………………………………………………………………………………………..

Paragraph-1 Enthalpy of reaction (dH) is measured when the reaction is carried out at constant pressure. When a bomb calorimeter is used, the volume does not change. The amount of energy measured is the internal energy dE. To convert dE into dH, we use the defined relationship, dH = dE + d(P V).The changes in pressure and volume (P V) work can be evaluated by the application of ideal gas law, d (P V) = dn R T, where dn is the total number of moles of gas of product Σn(products) substracting the total number of moles of gas of reactants, Σn(reactants). dn = Σn(products) - Σn(reactants) 15. Which of the following statements is false? (A)The value of q is positive when heat flows into a system from the surroundings. (B)Heat flows from the system into the surroundings in an exothermic process. ©In an endothermic process heat flows from the surroundings into the system. (D) Enthalpy is a state property. 16. Heat capacity is defined as (A) 4.18 cal/g °C. (B)4.18 J/g °C. ©the amount of heat necessary to raise the temperature of a system by 1°C. (D) the amount of heat necessary to raise the temperature of 1 gram of a substance by 1°C. 17. In which of the reactions below is ΔH not an enthalpy of formation? (A). C(s) + O2(g) → CO2(g) (B) 2C(s) + O2(g) → 2CO(g) © Ca(s) + 1/2O2(g) → CaO(s) (D) both b and c are not enthalpy of formation reactions

Paragraph-2 The standard entropy of a substance is its entropy at 1 atm pressure. The values found in tables are normally those for 298K, and are expressed in units of J K–1 mol–1. The table below shows some typical values for gaseous substances.



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He 126 H2 131 CH4 186 Ne 146 N2 192 H2O(g) 187 Ar 155 CO 197 CO2 213 Kr 164 F2 203 C2H6 229 Xe 170 O2 205 n -C3H8 270 Cl2 223 n -C4H10 310

Table 1: Standard entropies of some gases at 298K, J K–1 mol–1 • Although the standard internal energies and enthalpies of these substances would be zero, the entropies are not. This is because there is no absolute scale of energy, so we conventionally set the “energies of formation” of elements in their standard states to zero. Entropy, however, measures not energy itself, but its dispersal amongst the various quantum states available to accept it, and these exist even in pure elements. • It is apparent that entropies generally increase with molecular weight. For the noble gases, this is of course a direct reflection of the principle that translational quantum states are more closely packed in heavier molecules, allowing of them to be occupied. • The entropies of the diatomic and polyatomic molecules show the additional effects of rotational quantum levels.

C(diamond) C(graphite) Fe Pb Na S(rhombic) Si W 2.5 5.7 27.1 51.0 64.9 32.0 18.9 33.5

Table 2: Entropies of some solid elements at 298 K, J K–1 mol–1 The entropies of the solid elements are strongly influenced by the manner in which the atoms are bound to one another. The contrast between diamond and graphite is particularly striking; graphite, which is built up of loosely-bound stacks of hexagonal sheets, appears to be more than twice as good at soaking up thermal energy as diamond, in which the carbon atoms are tightly locked into a three-dimensional lattice, thus affording them less opportunity to vibrate around their equilibrium positions. Looking at all the examples in the above table, you will note a general inverse correlation between the hardness of a solid and its entropy. Thus sodium, which can be cut with a knife, has almost twice the entropy of iron; the much greater entropy of lead reflects both its high atomic weight and the relative softness of this metal. These trends are consistent with the oft-expressed principle that the more “disordered” a substance, the greater its entropy. 18. A Reaction which has a positive value of ΔS: (A)Will never proceed (B)Will always proceed ©Absorbs heat from the surroundings (D)Will proceed as long as ΔH is not too positive 19.The temperature at which the free energy change of a reaction is zero: (A) Can not be predicted (B)Is equal to ΔS/ΔH ©Is equal to ΔH/ΔS (D)Is equal to ΔG/ΔH 20. When solid Ba(OH)2 is added to water, it immediately dissolves and lowers the temperature of the solution. Which statement is true for this process? (A). ΔH > 0 and ΔS > 0 (B). ΔH > 0 and ΔS < 0 ©. ΔH < 0 and ΔS > 0 (D). ΔH < 0 and ΔS < 0

Paragraph-3 . YOU NEED TO REMEMBER THAT THE HEAT OF FORMATION OF ANY FREE ELEMENT IN ITS NORMAL PHYSICAL STATE AT ROOM TEMPERATURE IS "0 kJ/mole (whether it is listed in the table or not). YOU NEED TO KNOW A DEFINITION OF ΔHf

o , or enthalpy of formation. It is defined as the amount of heat absorbed or released when ONE mole of compound is formed FROM ITS CONSTITUENT ELEMENTS. By combining the enthalpy of formation values of different compounds, we can find how much energy is absorbed or released in a chemical reaction. The equation we will use is ΔHrxn = Σ nΔHf

o Products – Σ nΔHfo Reactants

Notice that what we are finding now is CHANGE OF ENTHALPY of the REACTION If the ΔHrxn is negative, it means that number of kiloJoules of energy is released in the balanced equation, and we say that the reaction is EXOTHERMIC. If the ΔHrxn is positive, it means that number of kiloJoules of energy must be added to the balanced equation in order to make it react and we say that the reaction is ENDOTHERMIC.



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21. The enthalpy of formation of H2O (l) is 285.83 kJ mol–1 and enthalpy of neutralisation of a strong acid and a

strong base is –55.84 kJ mol–1. The enthalpy of formation of OH– ions is (A) 341.67 kJ mol–1 (B) 229.99 kJ mol–1 (C) –229.99 kJ mol–1 (D) –341.67 kJ mol–1 22. ΔHºf for the reaction,

Ag+(aq) + Cl–(aq) ⎯→ AgCl(s) at 25ºC. Given ΔHºf (Ag+, aq) = 105.58 kJ mol–1, ΔHºf (Cl–, aq) = –167.16 kJ mol–1 and

ΔHºf (AgCl, s) = –127.07 kJ mol–1.

(A) –65.49 kJ mol–1 (B) 65.49 kJ mol–1 (C) 188.65 kJ mol–1 (D) –188.65 kJ mol–1 23. The change in state is HCl(g) + aq ⎯→ H+(aq) + Cl–(aq) Given ΔHºf (HCl, g) = –92.31 kJ mol–1 and ΔHºf (Cl–, aq) = –167.16 kJ mol–1, then The enthalpy change when one mol of HCl(g) is dissolved in a very large amount of water at 25ºC is – (A) 259.47 kJ mol–1 (B) 74.85 kJ mol–1 (C) –259.47 kJ mol–1 (D) –74.85 kJ mol–1

iit part test 1 thermodynamics by S K Sinha Kota by S.K.sinha See Chemistry Animations at...

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