Answer:
See explanation below.
Explanation:
We can obtain the Gibb's free energy from the formula;
∆G= ∆H - T∆S
Where;
∆G = change in free energy= the unknown
∆H= change in enthalpy = 3352 kJ
∆S= change in entropy of the solution= 625.1 J/K
T= absolute temperature = 298 K
Substituting values;
∆G= 3352 ×10^3 J - (298 K × 625.1 J/K)
∆G= 3352 ×10^3 J - 186279.8
∆G= 3.16 × 10^6 J
At 5975K,
∆G= ∆H - T∆S
∆G= 3352 ×10^3 J - (5975K × 625.1 J/K)
∆G= 3.352 ×10^6 J - 3.735 × 10^6
∆G= -3.83×10^5 J
At equilibrium, ∆G=0, Teq is given by;
0= 3352 ×10^3 J - (Teq × 625.1 J/K)
0= 3352 ×10^3 - 625.1Teq
625.1Teq = 3352 ×10^3
Teq= 3352 ×10^3/625.1
Teq= 5362.3 K
Standard Gibb's free energy at 298 & 5975K is 3.16×10⁶J and -3.83×10⁵J respectively, and temperature at equilibrium state is 5362.3 K.
Standard Gibb's free energy of any reaction will be calculated as:
∆G° = ∆H° - T∆S°, where
∆G° = change in free energy = to find?
∆H° = change in enthalpy = 3352 kJ = 3,352 × 10³J
∆S° = change in entropy = 625.1 J/K
T = absolute temperature = 298 K
On putting all these values on the above equation, we get
∆G° = (3,352 × 10³) - (298 × 625.1) = 3.16 × 10⁶J
Gibb's free energy at 5,975 K temperature:
∆G° = (3,352 × 10³) - (5,975 × 625.1) = -3.83 × 10⁵J
At the equilibrium state value of Gibb's free energy is zero, so from the equation and given data we can calculate the value of temperature as:
0 = (3352 × 10³)J - (T × 625.1 J/K) = 5362.3 K
Hence required values are 3.16 × 10⁶J, -3.83 × 10⁵J and 5362.3 K.
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https://brainly.com/question/9908454
