Thermodynamics

JEE Main 2025 (Online) 23rd January Evening ShiftNumericalMedium📈 Topic in 18% papers+4

Ques. 1JEE Main 2025, 23 January (Evening Shift)

The bond dissociation enthalpy of X2 H bond calculated from the given data is kJ mol-1. (Nearest integer) M+X-(s) M+(g)+X-(g) H lattice=800 kJ mol-1 M(s) M(g) H sub=100 kJ mol-1 M(g) M+(g)+e-(g) H i=500 kJ mol-1 X(g)+e-(g) X-(g) H eg=-300 kJ mol-1 M(s)+ 12X2(g) M+X-(s) H f=-400 kJ mol-1 [Given : M+X- is a pure ionic compound and X forms a diatomic molecule X2 in gaseous state]

Answer

200

Solution

Born–Haber: the long route must end at the same level as the direct route Hf.

[figure]

-400=100+500+ 12 Hbond-300-800

⇒ Hbond=200 kJ/mol

Answer: 200

Keep practising — Thermodynamics

Ques. 2JEE Main 2025, 22 January (Morning Shift)

A liquid when kept inside a thermally insulated closed vessel at 25 was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ?

U = 0, q < 0, w > 0
U > 0, q = 0, w > 0
U = 0, q = 0, w = 0
U < 0, q = 0, w > 0

Ques. 3JEE Main 2025, 22 January (Evening Shift)

Match List - I with List - II. Choose the correct answer from the options given below :

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

Ques. 4JEE Main 2025, 22 January (Evening Shift)

Consider the following cases of standard enthalpy of reaction ( Hr in kJ mol-1) C2H6(g)+ 72O2(g) 2CO2(g)+3H2O(l) H1 =-1550 C(graphite)+O2(g) CO2(g) H2 =-393.5 H2(g)+ 12O2(g) H2O(l) H3 =-286 The magnitude of H f C2H6(g) is kJ mol-1 (Nearest integer).

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