H2(g) + Br2(g) → 2HBr(g)
The above reaction is...
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
Correct answer is B
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.
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