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JEE Exam  >  JEE Tests  >  Daily Test for JEE Preparation  >  Chemistry Unit Test : Thermodynamics (June 15) - JEE MCQ

Chemistry Unit Test : Thermodynamics (June 15) - JEE MCQ


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15 Questions MCQ Test Daily Test for JEE Preparation - Chemistry Unit Test : Thermodynamics (June 15)

Chemistry Unit Test : Thermodynamics (June 15) for JEE 2024 is part of Daily Test for JEE Preparation preparation. The Chemistry Unit Test : Thermodynamics (June 15) questions and answers have been prepared according to the JEE exam syllabus.The Chemistry Unit Test : Thermodynamics (June 15) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Chemistry Unit Test : Thermodynamics (June 15) below.
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Chemistry Unit Test : Thermodynamics (June 15) - Question 1
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The standard heat of formation () of ethane (in kJ/mol), if the heat of combustion of ethane, hydrogen and graphite is -1560, -393.5 and -286 kJ/mol, respectively, is ______.(Answer up to 1 decimal place)


Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 1

(1) C + O2  CO2,  = -286 kJ/mol
(2) H2 + O2  H2O,  = -393.5 kJ/mol
(3) C2H6 + O2  2CO2 + 3H2O,  = -1560 kJ/mol
Calculation: 2 × (1) + 3 × (2) - (3)
2C + 3H2  C2H6
(C2H6) = 2(-286) + 3(-393.5) + 1560 = -192.5 kJ/mol

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Chemistry Unit Test : Thermodynamics (June 15) - Question 2
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4.0 L of an ideal gas is allowed to expand isothermally into vacuum until the total volume is 20 L. The amount of heat absorbed in this expansion is _____________ L atm. (In integer)


Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 2

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Chemistry Unit Test : Thermodynamics (June 15) - Question 3
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While performing a thermodynamics experiment, a student made the following observations:
HCl + NaOH →NaCl + H2O ΔH = -57.3 kJ mol-1
CH3COOH + NaOH →CH3COONa + H2O ΔH = -55.3 kJ mol-1
The enthalpy of ionization of CH3COOH as calculated by the student is ________ kJ mol-1. (Nearest integer)


Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 3

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Chemistry Unit Test : Thermodynamics (June 15) - Question 4
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A cylinder containing an ideal gas (0.1 mol of 1.0 dm3) is in thermal equilibrium with a large volume of 0.5 molal aqueous solution of ethylene glycol at its freezing point. If the stoppers S1 and S2 (as shown in the figure) are suddenly withdrawn, the volume of the gas in litres after equilibrium is achieved will be ________.(Answer up to 2 decimal places)

(Given, Kf (water) = 2.0 K kg mol-1, R = 0.08 dm3 atm K-1 mol-1)

Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 4

For aqueous solution,
 = Kf.m = 2 × 0.5
∴ Temperature of the solution = -1°C = 272 K
∴ Final volume of the ideal gas = 
= 
≈ 2.18 L

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Chemistry Unit Test : Thermodynamics (June 15) - Question 5
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Among the following, the number of state variables is ______.(In integer)

Internal energy (U)
Volume (V)
Heat (q)
Enthalpy (H)

Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 5

State variables → Volume, Enthalpy, Internal Energy
Path variable → Heat

Chemistry Unit Test : Thermodynamics (June 15) - Question 6
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Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol-1 and 4 kJ mol-1, respectively. The hydration enthalpy of NaCl is
Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 6

Chemistry Unit Test : Thermodynamics (June 15) - Question 7
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According to the following diagram, A reduces BO2 when the temperature is
Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 7

Reduction of BO2 using A:

From the given diagram, it is clear that G° of the required reaction will be negative when T > 1400°C.

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Chemistry Unit Test : Thermodynamics (June 15) - Question 8
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The molar heat capacity for an ideal gas at constant pressure is 20.785 J K-1mol-1. The change in internal energy is 5,000 J upon heating it from 300 K to 500 K. The number of moles of the gas at constant volume is ______. [Nearest Integer]
(Given: R = 8.314 J K-1 mol-1)

Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 8

And we know that

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Chemistry Unit Test : Thermodynamics (June 15) - Question 9
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At 300 K, a sample of 3.0 g of gas A occupies the same volume as 0.2 g of hydrogen at 200 K at the same pressure. The molar mass of gas A is ______ g mol-1. (Nearest integer)
Assume that the behaviour of gases as ideal. (Given: The molar mass of hydrogen (H2) gas is 2.0 g mol-1)

Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 9

V1, Volume of 0.2 g H2 at 200 K = 
V2, Volume of 3.0 g of gas A at 300 K = 
V1 = V2 (Given)

 M = 45 g mol-1

Chemistry Unit Test : Thermodynamics (June 15) - Question 10
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If enthalpy of atomisation for Br2(l) is x kJ/mol and bond enthalpy for Br2 is y kJ/mol, the relation between them
Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 10

Chemistry Unit Test : Thermodynamics (June 15) - Question 11
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At 25°C and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) is -3268 kJ mol-1 and -1300 kJ mol-1, respectively. The change in enthalpy for the reaction 3C2H2(g) →+324 kJ mol-1 C6H6(l) is
Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 11

I. 
II. 
III. 
Applying Hess's law of constant heat summation,
ΔH3 = 3 × ΔH2 - ΔH1
= 3 × (-1300) - (-3268)
= -632 kJ mol-1

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Chemistry Unit Test : Thermodynamics (June 15) - Question 12
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The standard entropy change for the reaction
4Fe(s) + 3O2(g) → 2Fe2O3(s) is -550 JK-1 at 298 K.
[Given: The standard enthalpy change for the reaction is -165 kJ mol-1]. The temperature in K at which the reaction attains equilibrium is __________. (Nearest Integer)

Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 12

Chemistry Unit Test : Thermodynamics (June 15) - Question 13
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Which one of the following is correct for the adsorption of a gas at a given temperature on a solid surface?
Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 13

(i) Adsorption of gas at metal surface is an exothermic process, so ΔH < 0.
(ii) The adsorption of gas on metal surface reduces the free movement of gas molecules, thus restricting its randomness, so ΔS < 0.

Chemistry Unit Test : Thermodynamics (June 15) - Question 14
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At 25°C and 1 atm pressure, the enthalpies of combustion are as given below:

The enthalpy of formation of ethane is
Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 14

2C(graphite) + 3H2(g) → C2H6(g)
Hr = +1560 + 2(−394) + 3(−286)
= -86.0 kJ mol-1
Enthalpy of formation of C2H6(g) = -86.0 kJ mol-1

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Chemistry Unit Test : Thermodynamics (June 15) - Question 15
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For the reaction:
A(l) →2B(g),
ΔU = 2.1 kcal, ΔS = 20 cal K-1 at 300 K.

Hence, ΔG in kcal is _________.(Answer up to one decimal place)

Detailed Solution for Chemistry Unit Test : Thermodynamics (June 15) - Question 15

We know:
ΔG = ΔH - TΔS
ΔH = ΔU + 2RT
Thus, we have:
ΔG = ΔU + 2RT - TΔS
On putting the given values, we get:
ΔG = 2.1 + 2 × 2 × 300 × 10-3 - 300 × 20 × 10-3
ΔG = 2.1 + 4 × 300 × 10-3 - 300 × 20 × 10-3
ΔG = 2.1 + 1200 × 10-3 - 6000 × 10-3
ΔG = 2.1 + 1.2 - 6
ΔG = 3.3 - 6
ΔG = -2.7 kcal

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