Yes, and that means dH must be large(negative) eough to off set dS.
BTW, a little urgent is like being a little pregnant.
4Fe+3O2=2Fe2O3
the entropy change is -549.45JK^-1mol^-1 at 298K. Though it has negative entropy change in the reaction is spotaneous.why?
(Delta r H theta ∆rH°=-1648×10^3 J ml^-1.
Is it because ∆G is negative????
BTW, a little urgent is like being a little pregnant.
The relationship between ∆G, ∆H, and ∆S is given by the equation:
∆G = ∆H - T∆S,
where ∆H is the enthalpy change, T is the temperature (in Kelvin), and ∆S is the entropy change.
Even though the entropy change is negative (∆S < 0), it can still be compensated by the enthalpy change (∆H) at a given temperature (T). If the enthalpy change is significantly negative (∆H < 0), it can outweigh the negative entropy change, resulting in a negative ∆G, making the reaction spontaneous.
Therefore, in the case of the oxidation of iron, even though the entropy change (∆S) is negative, the large negative enthalpy change (∆H) dominates, resulting in a negative ∆G and making the reaction spontaneous.
∆G = ∆H - T∆S
where T is the temperature in Kelvin. If ∆G is negative, the reaction is spontaneous under the given conditions.
In this case, you have mentioned that the entropy change (∆S) for the oxidation of iron is -549.45 J K^-1 mol^-1. At a temperature of 298 K, we can calculate the value of ∆G using the equation mentioned above:
∆G = ∆H - T∆S
∆G = (-1648 × 10^3 J mol^-1) - (298 K) * (-549.45 J K^-1 mol^-1)
∆G = -1648 × 10^3 J mol^-1 + 163,702.1 J mol^-1
∆G = -1485 × 10^3 J mol^-1
Since the value of ∆G is negative (-1485 × 10^3 J mol^-1), it indicates that the reaction is spontaneous under standard conditions (298 K, 1 atm pressure, 1 M concentrations). So, yes, in this case, the negative value of ∆G indicates that the reaction is spontaneous.