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
Explanation
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Discussions (11)
The ans is B. They kinda made a mockery of us, knowing we wont realise the kj and j.
Remember
∆G = ∆H - ∆S×T
= -72000- ( -106×298)
=- 40.4kj
∆G is negative therefore the reactiin is crazily spontanoeus.
In exothermic reaction, negative enthalpy means the reaction is spontaneous while positive mean the reaction is non spontaneous.
In this question reported, after I finished my calculations I found that the free energy is negative which means the reaction is spontaneous.
Here is my calculations:
∆G=∆H-T∆S
Where;
∆G=?
∆H= -72KJ/mol
T=25°C =(25+273)K =298K
∆S= -106J/mol = -0.106KJ/mol/K
Substituting the value in the equation above; we have:
∆G= -72-(298× -0.106) = -72 +31.588
∆G= -40.412KJ/mol
for a reaction to be proceed spontaneously entropy change must be positive and in this case i think we are seeing the reverse consequently it will not proceed
Definitely the answer is B , because after carefully resolving it , free energy is -40.412kjmol. Since it is negative, the reaction is spontaneous as it is one of the condition for spontaneity of a reaction.
So does this mean that the negative entropy doesn't matter bcos I think that for a reaction to be spontaneous, entropy shud be positive and free energy negative
