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The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the standard molar Gibbs energy of formation of X(g) is 5.61 kJ·mol−1 at 2000. K and −52.80 kJ·mol−1 at 3000. K. Determine the value of K (the thermodynamic equilibrium constant) at each temperature. K at 2000. K= K at 3000. K= Assuming that ΔH∘rxn is independent of temperature, determine the value of ΔH∘rxn from this data. ΔH∘rxn=

Answer :

Answer:

The equilibrium constant at 2000 K is 0.7139

The equilibrium constant at 3000 K is 8.306

ΔH = 122.2 kJ/mol

Explanation:

Step 1: Data given

the standard molar Gibbs energy of formation of X(g) is 5.61 kJ/mol at 2000 K

the standard molar Gibbs energy of formation of X(g) is -52.80 kJ/mol at 3000 K

Step 2: The equation

1/2X2(g)⟶X(g)

Step 3: Determine K at 2000 K

ΔG = -RT ln K

⇒R = 8.314 J/mol *K

⇒T = 2000 K

⇒K is the equilibrium constant

5610 J/mol = -8.314 J/molK * 2000 * ln K

ln K = -0.337

K = e^-0.337

K = 0.7139

The equilibrium constant at 2000 K is 0.7139

Step 4: Determine K at 3000 K

ΔG = -RT ln K

⇒R = 8.314 J/mol *K

⇒T = 3000 K

⇒K is the equilibrium constant

-52800 J/mol = -8.314 J/molK * 3000 * ln K

ln K = 2.117

K = e^2.117

K = 8.306

The equilibrium constant at 3000 K is 8.306

Step 5: Determine the value of ΔH∘rxn

ln K2/K1 = -ΔH/r * (1/T2 - 1/T1)

ln 8.306 /0.713 = -ΔH/8.314 * (1/3000 - 1/2000)

2.455 = -ΔH/8.314 * (3.33*10^-4 - 0.0005)

2.455 = -ΔH/8.314 * (-1.67*10^-4)

-14700= -ΔH/8.314

-ΔH = -122200 J/mol

ΔH = 122.2 kJ/mol

When The constant at 2000 K is 0.7139

Then The constant at 3000 K is 8.306

ΔH = 122.2 kJ/mol

What is Equilibrium?

Step 1: Data is given

When the quality molar Gibbs effectiveness of the formation of X(g) is 5.61 kJ/mol at 2000 K

When the quality molar Gibbs effectiveness of the formation of X(g) is -52.80 kJ/mol at 3000 K

Step 2: The equation is:

[tex]1/2X2(g)⟶X(g)[/tex]

Step 3: Then Determine K at 2000 K

ΔG = -RT ln K

[tex]⇒R = 8.314 J/mol *K[/tex]

⇒[tex]T = 2000 K[/tex]

⇒ at that time K is that the constant

Then 5610 J/mol = [tex]-8.314 J/molK * 2000 * ln K[/tex]

ln K is = [tex]-0.337[/tex]

K is =[tex]e^-0.337[/tex]

K is = [tex]0.7139[/tex]

When The constant at 2000 K is 0.7139

Step 4: Then Determine K at 3000 K

ΔG = -RT ln K

⇒[tex]R = 8.314 J/mol *K[/tex]

⇒[tex]T = 3000 K[/tex]

⇒K is that the constant

[tex]-52800 J/mol = -8.314 J/mol K * 3000 * ln K[/tex]

ln [tex]K = 2.117[/tex]

K = e^2.117

K = 8.306

The constant at 3000 K is 8.306

Step 5: at the moment Determine the worth of ΔH∘rxn

ln K2/K1 = -ΔH/r * (1/T2 - 1/T1)

ln 8.306 /0.713 = -Δ[tex]H/8.314 * (1/3000 - 1/2000)[/tex]

2.455 = -ΔH/8.314 * (3.33*10^-4 - 0.0005)

2.455 = -ΔH/8.314 * [tex](-1.67*10^-4)[/tex]

-14700= -ΔH/8.314

-ΔH = [tex]-122200 J/mol[/tex]

Then ΔH = [tex]122.2 kJ/mol[/tex]

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