Refer to BITSAT Chemistry Thermodynamics MCQs provided below. BITSAT Full Syllabus Chemistry MCQs with answers available in Pdf for free download. The MCQ Questions for Full Syllabus Chemistry with answers have been prepared as per the latest syllabus, BITSAT books and examination pattern suggested in Full Syllabus by BITSAT, NCERT and KVS. Multiple Choice Questions for Thermodynamics are an important part of exams for Full Syllabus Chemistry and if practiced properly can help you to get higher marks. Refer to more Chapter-wise MCQs for BITSAT Full Syllabus Chemistry and also download more latest study material for all subjects
MCQ for Full Syllabus Chemistry Thermodynamics
Full Syllabus Chemistry students should refer to the following multiple-choice questions with answers for Thermodynamics in Full Syllabus. These MCQ questions with answers for Full Syllabus Chemistry will come in exams and help you to score good marks
Thermodynamics MCQ Questions Full Syllabus Chemistry with Answers
Question: For the reaction Which one of the statement is correct at constant T and P ?
a) ΔH=ΔE
b) ΔH< ΔE
c) ΔH>ΔE
d) ΔH is independent of physical state of the reactants
Answer: ΔH< ΔE
Question: For a given reaction, ΔH = 35.5 kJ mol-1 and ΔS = 83.6 JK-1 mol-1 1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with tempearature)
a) T > 425 K
b) All temperatures
c) T > 298 K
d) T < 425 K
Answer: T > 425 K
Question: For a particular reversible reaction at temperature T, ΔH and ΔS were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when
a) Te > T
b) T > Te
c) Te is 5 times T
d) T = Te
Answer: T > Te
Question: Given
Based on data provided, the value of electron gain enthalpy of fluorine would be :
a) – 300 kJ mol–1
b) – 350 kJ mol–1
c) – 328 kJ mol–1
d) – 228 kJ mol–1
Answer: – 328 kJ mol–1
Question: Which law of the thermodynamics helps in calculating the absolute entropies of various substances at different temperatures?
a) First law
b) Second law
c) Third law
d) Zeroth law
Answer: Third law
Question: A spontaneous reaction is impossible if
a) Both ΔH and ΔS are negative
b) Both ΔH and ΔS are positive
c) ΔH is negative and ΔS is positive
d) ΔH is positive and ΔS is negative
Answer: ΔH is positive and ΔS is negative
Question: If the enthalpy of vaporization of water is 186.5 kJmol–1, the entropy if its vaporization will be :
a) 0.5 JK–1mol–1
b) 1.0 JK–1mol–1
c) 1.5 JK–1mol–1
d) 2.0 JK–1mol–1
Answer: 0.5 JK–1mol–1
Question: The heats of neutralisation of CH3COOH, HCOOH, HCN and H2S are – 13.2, – 13.4, – 2.9 and – 3.8 kCal per equivalent respectively. Arrange the acids in increasing order of acidic strength.
a) HCOOH > CH3COOH > H2S > HCN
b) CH3COOH > HCOOH > H2S > HCN
c) H2S > HCOOH > CH3COOH > HCN
d) HCOOH > H2S > CH3COOH > HCN
Answer: HCOOH > CH3COOH > H2S > HCN
Question: For determining the spontaneity of a process which of the following is considered ?
a) ΔS system
b) ΔS surroundings
c) ΔS system + ΔS surroundings
d) ΔS system – ΔS surroundings
Answer: ΔS system + ΔS surroundings
Question: The enthalpy of combustion of 2 moles of benzene at 27°C differs from the value determined in bomb calorimeter by
a) – 2.494 kJ
b) 2.494 kJ
c) – 7.483 kJ
d) 7.483 kJ
Answer: – 7.483 kJ
Question: In which of the following reactions, standard entropy change (ΔS°) is positive and standard Gibb’s energy change (ΔG°) decreases sharply with increasing temperature ?
a)
b)
c)
d)
Answer:
Question: Bond enthalpies of H2 , X2 and HX are in the ratio 2 : 1 : 2. If enthalpy of formation of HX is – 50 kJ mol–1, the bond enthalpy of X2 is
a) 100 kJ mol–1
b) 300 kJ mol–1
c) 200 kJ mol–1
d) 400 kJ mol–1
Answer: 100 kJ mol–1
Question: What is the free energy change, ΔG , when 1.0 mole of water at 100º C and 1 atm pressure is converted into steam at 100°C and 1 atm. pressure?
a) 540 cal
b) –9800 cal
c) 9800 cal
d) 0 cal
Answer: 0 cal
Question: For the reaction : Which one of the following is correct regarding ΔH
a) ΔH = ΔE + 2RT
b) ΔH = ΔE – 2RT
c) ΔH = ΔE + RT
d) ΔH = ΔE – RT
Answer: ΔH = ΔE – 2RT
Question: One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1L to 10 L. The ΔE for this process is (R = 2 cal mol–1 K–1)
a) 163.7 cal
b) zero
c) 1381.1 cal
d) 9 lit atom
Answer: zero
Question: At 25°C and 1 bar which of the following has a non-zero ΔH°f ?
a) Br2(l)
b) C (graphite)
c) O3(g)
d) I2(s)
Answer: O3(g)
Question: For which of the process, ΔS is negative?
a) H2 (g)→ 2H(g)
b) N2 (g )(1atm)→ N2 (g)(8atm)
c) 2SO3 (g)→ 2SO2 (g) +O2 (g)
d) C(diamond) → C(graphite)
Answer: N2 (g )(1atm)→ N2 (g)(8atm)
Question: Ka for CH3COOH at 25°C is 1.754 × 10–5 . At50°C Ka is 1.633 × 10–5. What are ΔH° and ΔS° for the ionisation of CH3COOH ?
a) – 1.55 kJmol-1, – 96.44 Jmol -1K-1
b) – 2.55 kJmol-1, – 106.44 Jmol -1K-1
c) 1.55 kJmol-1, 96.44 Jmol -1K-1
d) – 1.55 kJmol-1, 96.44 Jmol -1K-1
Answer: – 1.55 kJmol-1, – 96.44 Jmol -1K-1
Question: Temperature of 5 moles of a gas is decreased by 2K at constant pressure. Indicate the correct statement
a) Work done by gas is = 5 R
b) Work done over the gas is = 10 R
c) Work done by the gas = 10 R
d) Work done = 0
Answer: Work done over the gas is = 10 R
Question: Compounds with high heat of formation are less stable because
a) It is difficult to synthesize them
b) Energy rich state leads to instability
c) High temperature is required to synthesize them
d) Molecules of such compunds are distorted
Answer: Energy rich state leads to instability
Question: The energy that opposes dissolution of a solvent is
a) Hydration energy
b) Lattice energy
c) Internal energy
d) Bond energy
Answer: Lattice energy
Question: The energy that opposes dissolution of a solvent is
a) Hydration energy
b) Lattice energy
c) Internal energy
d) Bond energy
Answer: Lattice energy
Question: In the reaction ΔH = 40 kJ; ΔH represents :
a) Heat of formation
b) Heat of combustion
c) Heat of neutralisation
d) Heat of reaction
Answer: Heat of reaction
Question: The energy that opposes dissolution of a solvent is
a) Hydration energy
b) Lattice energy
c) Internal energy
d) Bond energy
Answer: Lattice energy
Question: Which law of the thermodynamics helps in calculating the absolute entropies of various substances at different temperatures?
a) First law
b) Second law
c) Third law
d) Zeroth law
Answer: Third law
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BITSAT Full Syllabus Chemistry Thermodynamics MCQs
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BITSAT MCQs Chemistry Full Syllabus Thermodynamics
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