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Standard Gibb's free energy at 298K and 5975K is 3.16×10⁶J and -3.83×10⁵J respectively, and temperature at equilibrium state is 5362.3 K.
What is the standard Gibb's free energy?
The complete question is in the brainly link at the end. We can say that;
Standard Gibb's free energy of any reaction is gotten from the formula;
∆G° = ∆H° - T∆S°
where;
∆G° is change in free energy
∆H° is change in enthalpy = 3352 kJ = 3352 × 10³J
∆S° is change in entropy = 625.1 J/K
T is absolute temperature = 298 K
Plugging in the relevant values into the gibbs energy equation gives;
∆G° = (3352 × 10³) - (298 × 625.1)
∆G° = 3.16 × 10⁶J
Gibb's free energy at temperature of 5,975 K is;
∆G° = (3,352 × 10³) - (5,975 × 625.1)
∆G° = -3.83 × 10⁵J
At the equilibrium state, the value of Gibb's free energy is zero. Thus we can calculate the temperature as;
0 = (3352 × 10³)J - (T × 625.1 J/K)
Solving for T gives;
T = 5362.3 K
Read more about Standard gibbs free energy at; https://brainly.com/question/17205875
