<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
When an oxygen atom is attached to a carbon atom, the carbon atom becomes reduced.
Answer:
Explanation:
Stoichiometry is based on the <u>conservation of mass</u>.
The added KI does not have any impact
The reaction invovles Titration of vitaminc ( Ascorbic acid)
ascorbic acid + I₂ → 2 I⁻ + dehydroascorbic acid
the excess iodine is free reacts with the starch indicator, forming the blue-black starch-iodine complex.
This is the endpoint of the titration. since alreay excess KI is added ( the source of Iodine), it does not have an influence.
Answer B
Hope this helps!
You have 0.50 mol of NH3 and 0.20 mol of NH4+ to start (NH4Cl dissolves completely), given the molarity and 1.0 L solution.
30.0 mL of 1.0 M HCl is 0.0300 mol of HCl. This will react with the NH3 to produced 0.030 mol of NH4+.
You now have 0.47 mol NH3 and 0.23 mol NH4+. Now use the Henderson-Hasselbach equation to calculate your pH. The equation says to use concentration of acid and base, but you can just use the moles of them because it doesn’t make a difference.
pH = pKa + log(base/acid)
pKa = 14 - pKb = 14 - 4.75 = 9.25
pH = 9.25 + log(0.47/0.23) = 9.56
