First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
Sodium is very reactive but it’s a metal, and the problem asks specifically for a non-metal.
Silicone is technically reactive, but not super reactive.
Argon is a nonmetal, however it is an inert gas. It doesn’t react with anything.
We’re left with Chlorine, which is a non-metal in group 7, a highly reactive group, on the periodic table.
<u>Answer:</u> The standard free energy change of formation of 
is 92.094 kJ/mol
<u>Explanation:</u>
We are given:
Relation between standard Gibbs free energy and equilibrium constant follows:
where,
= standard Gibbs free energy = ?
R = Gas constant = 
T = temperature = ![25^oC=[273+25]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5DK%3D298K)
K = equilibrium constant or solubility product = 
Putting values in above equation, we get:
For the given chemical equation:
The equation used to calculate Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the Gibbs free energy change of the above reaction is:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag%5E%2B%28aq.%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag_2S%28s%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol](https://tex.z-dn.net/?f=285.794%3D%5B%282%5Ctimes%2077.1%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%28-39.5%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%3D92.094J%2Fmol)
Hence, the standard free energy change of formation of is 92.094 kJ/mol
explain the question your asking
Answer:
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