<u>Answer:</u> The equilibrium constant for this reaction is 
<u>Explanation:</u>
The equation used to calculate standard Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the standard Gibbs free change of the above reaction is:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_{(NH_3(g))})]-[(1\times \Delta G^o_{(N_2)})+(3\times \Delta G^o_{(H_2)})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_%7B%28NH_3%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28N_2%29%7D%29%2B%283%5Ctimes%20%5CDelta%20G%5Eo_%7B%28H_2%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-16.45%29%29%5D-%5B%281%5Ctimes%20%280%29%29%2B%283%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D-32.9kJ%2Fmol)
To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:
where,
= standard Gibbs free energy = -32.9 kJ/mol = -35900 J/mol (Conversion factor: 1 kJ = 1000 J )
R = Gas constant = 8.314 J/K mol
T = temperature = ![25^oC=[273+25]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5DK%3D298K)
= equilibrium constant at 25°C = ?
Putting values in above equation, we get:
Hence, the equilibrium constant for this reaction is
Double the work done by the object.Hope this helps
The electronic configuration for vanadium (V) in the periodic table is as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3 (option D).
<h3>What is electronic configuration?</h3>
Electronic configuration is the the arrangement of electrons in an atom, molecule, or other physical structure like a crystal.
Vanadium is the 23rd element on the periodic table and has chemical symbol V with atomic number 23. It is a transition metal, used in the production of special steels.
This suggests that the electronic configuration of Vanadium will be written as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3
Therefore, the electronic configuration for vanadium (V) in the periodic table is as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3.
Learn more about electronic configuration at: brainly.com/question/14283892
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