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?
A
not proceed spontaneously at the given
B
proceed spontaneously at the given temperature
C
proceed in the reverse direction at the given temperature
D
proceed spontenously at lower temperature
Verified Correct Answer: B.
Step-by-Step Mathematical Solution
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.
Source: WAEC / JAMB / NECO Classroom Archive (Record #3735)