H2(g) + Br2(g) → 2HBr(g)
The above reaction is carried out at 25oC. 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
The correct answer is: B
Explanation
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 as ∆G to be negative.