JEE Mains Model Paper 13

Ef = 320 KJ/mole , Eb = 400 KJ/mole
Given X ⇌ YΔH = - 80 KJ/mole ;

Eb

Ef

=

5

4

;
We know that Ef –Eb = ΔH , Ef – Eb = - 80,
Ef -

5

4

Ef = -80 (∵ Eb =

5

4

Ef)
Ef (1 -

5

4

:

4−5

4

=

1

4

) = 80 ; ∴ Ef = 320
∴ Ef = 320kJ/mole ; Eb = 400 kJ/mole