The equation represents the decomposition of a generic diatomic element in its standard state. 12x2(g)⟶x(g) 12x2(g)⟶x(g) assume that the standard molar gibbs energy of formation of x(g) is 5.24 kj·mol−15.24 kj·mol−1 at 2000. k and −65.59 kj·mol−1−65.59 kj·mol−1 at 3000. k. determine the value of kk (the thermodynamic equilibrium constant) at each temperature. assuming that δh∘rxn is independent of temperature, determine the value of δh∘rxn from this data.

The value of K at the different temperatures are 1.37 and 13.87

The equation of the reaction is; 12X2(g)⟶24X(g)

We have the following information;

ΔG =  5.24 kJ·mol−1 at  2000. K

ΔG = −65.59 kJ·mol−1 at 3000. K.

R = 8.314 J/K/mol

Given that;

ΔG = −RTlnK

K = e^-(ΔG/RT)

K = e^-(5.24 × 10^3J/mol/ 8.314 J/K/mol ×  2000)

K =1.37

Also;

ΔG = −RTlnK

K = e^-(ΔG/RT)

K = e^-(-65.59 × 10^3J/mol/ 8.314 J/K/mol ×  3000)

K =13.87

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At 2000 K,  K=0.72

At 3000 K,  K=14.7

Explanation:

Given that

ΔG = 5.24  KJ/mol ,T= 2000 K

ΔG =- 65.59  KJ/mol ,T= 3000 K

The Gibbs energy at temperature T is given as

ΔG =- RT ln K

R= Universal gas constant

R=8.314 J/mol.K

At T= 2000 K ,ΔG = 5.24  KJ/mol

ΔG =- RT ln K

5.24 x 1000 = 8.314 x 2000 ln K

K=0.72

At T= 32000 K ,ΔG = -65.59  KJ/mol

ΔG =- RT ln K

-65.59 x 1000 = -8.314 x 3000 ln K

K=14.7

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )