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?
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
Explication
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.