JEE Main 2022 (Online) 27th June Morning Shift | Thermodynamics Question 65 | Chemistry

1

JEE Main 2022 (Online) 27th June Morning Shift

MCQ (Single Correct Answer)

+4

-1

Match List-I with List-II.

	List - I		List - II
(A)	Spontaneous process	(I)	$$\Delta H < 0$$
(B)	Process with $$\Delta P = 0$$, $$\Delta T = 0$$	(II)	$$\Delta {G_{T,P}} < 0$$
(C)	$$\Delta {H_{reaction}}$$	(III)	Isothermal and isobaric process
(D)	Exothermic Process	(IV)	[Bond energies of molecules in reactants] $$ - $$ [Bond energies of product molecules

Choose the correct answer from the options given below :

A

(A) - (III), (B) - (II), (C) - (IV), (D) - (I)

B

(A) - (II), (B) - (III), (C) - (IV), (D) - (I)

C

(A) - (II), (B) - (III), (C) - (I), (D) - (IV)

D

(A) - (II), (B) - (I), (C) - (III), (D) - (IV)

2

JEE Main 2022 (Online) 25th June Evening Shift

MCQ (Single Correct Answer)

+4

-1

At 25$$^\circ$$C and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are $$-$$ 3268 kJ mol^$$-$$1 and $$-$$1300 kJ mol^$$-$$1, respectively. The change in enthalpy for the reaction 3 C₂H₂(g) $$\to$$ C₆H₆ (I), is :

A

+324 kJ mol^$$-$$1

B

+632 kJ mol^$$-$$1

C

$$-$$ 632 kJ mol^$$-$$1

D

$$-$$ 732 kJ mol^$$-$$1

3

JEE Main 2022 (Online) 24th June Evening Shift

MCQ (Single Correct Answer)

+4

-1

At 25$$^\circ$$C and 1 atm pressure, the enthalpies of combustion are as given below :

Substance	$${H_2}$$	C (graphite)	$${C_2}{H_6}(g)$$
$${{{\Delta _c}{H^\Theta }} \over {kJ\,mo{l^{ - 1}}}}$$	$$ - 286.0$$	$$ - 394.0$$	$$ - 1560.0$$

The enthalpy of formation of ethane is

A

+54.0 kJ mol^$$-$$1

B

$$-$$68.0 kJ mol^$$-$$1

C

$$-$$86.0 kJ mol^$$-$$1

D

+97.0 kJ mol^$$-$$1

4

JEE Main 2021 (Online) 31st August Evening Shift

MCQ (Single Correct Answer)

+4

-1

The incorrect expression among the following is :

A

$${{\Delta {G_{System}}} \over {\Delta {S_{Total}}}} = - T$$ (at constant P)

B

$$K = {{\Delta H^\circ - T\Delta S^\circ } \over {RT}}$$

C

$$K = {e^{ - \Delta G^\circ /RT}}$$

D

For isothermal process $${w_{reversible}} = - nRT\ln {{{V_f}} \over {{V_i}}}$$