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JAMB - Chemistry (2006 - No. 37)

H\(_2\)(g) + Br\(_2\)(g) → 2HBr(g)
The above reaction is carried out at 25\(^0\)C. ΔH is -72 kJ mol-1 and ΔS is - 106 J mol-1K-1, the reaction will?
not proceed spontaneously at the given
proceed spontaneously at the given temperature
proceed in the reverse direction at the given temperature
proceed spontenously at lower temperature

Објашњење

For a reaction to be spontaneous, ∆G = - ve; ∆H = - ve; ∆S = + ve

 Where ∆G = ∆H - T∆S

Given data; ∆G = ?

∆H = - 72KJ/mol, 

T = (25+273)K = 298K

∆S = -106J/mol or - 0.106KJ/mol/K

: ∆G = -72 -(298 * -  0.106)

= - 72 + 31.588

∆G = - 40.412KJ/mol

One of the condition for spontaneity of a reaction is ∆G to be negative.

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