<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
Answer:
a) pH will be 12.398
b) pH will be 4.82.
Explanation:
a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles
The total volume after addition of pure water = 0.780+0.01 = 0.79 L
The new concentration of /NaOH will be:
the [OH⁻] = 0.025
pOH = -log [OH⁻] = 1.602
pH = 14 -pOH = 12.398
b) The buffer has butanoic acid and butanoate ion.
i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:
![pH=pKa+log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
pKa=
ii) on addition of base the pH will increase.
I don't know your mc choices so I am going to go with "the number of valence electrons"
Oxidation number: the amount of electrons an atom gains, loses or shares when it forms chemical bonds. As for the valence electrons, those are the electrons that are responsible for bonds/ reactions.
Answer:
I dont understand what is the question you need help on-
Explanation:
