Answer : The value of 
for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
= enthalpy of reaction = ?
n = number of moles
= standard enthalpy of formation
Now put all the given values in this expression, we get:
![\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5B2mole%5Ctimes%20%28-242kJ%2Fmol%29%2B2mole%5Ctimes%20%28-296.8kJ%2Fmol%29%7D%5D-%5B2mole%5Ctimes%20%28-21kJ%2Fmol%29%2B3mole%5Ctimes%20%280kJ%2Fmol%29%5D)
conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction 
.
![\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28O_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of formation
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28189J%2FK.mol%29%2B2mole%5Ctimes%20%28248J%2FK.mol%29%7D%5D-%5B2mole%5Ctimes%20%28206J%2FK.mol%29%2B3mole%5Ctimes%20%28205J%2FK.mol%29%5D)
Now we have to calculate the Gibbs free energy of reaction 
.
As we know that,
At room temperature, the temperature is 500 K.
Therefore, the value of for the reaction is -959.1 kJ
15 grams of NH3 can be dissolved
<h3>Further explanation</h3>
Given
50 grams of water at 50°C
Required
mass of NH3
Solution
Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility
- 1. Temperature:
- 2. Surface area:
- 3. Solvent type:
- 4. Stirring process:
We can use solubility chart (attached) to determine the solubility of NH3 at 50°C
From the graph, we can see that the solubility of NH3 in 100 g of water at 50 C is 30 g
So that the solubility in 50 grams of water is:
= 50/100 x 30
= 15 grams
Answer:
b C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻
e HC₂H₃O₂ + H₂O ⇄ H₃O⁺ + C₂H₃O₂⁻
f ka<<1
g. Weak acid molecules and water molecules
Explanation:
The water molecule could act as a base and as an acid, a molecule that have this property is called as amphoteric.
b The salt NaC₂H₃O₂ is dissolved in water as Na⁺ and C₂H₃O₂⁻. The reaction of the anion with water is:
<em>C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻</em>
Where the C₂H₃O₂⁻ is the base and water is the acid.
e. The reaction of HC₂H₃O₂ (acid) with water (base), produce:
<em>HC₂H₃O₂ + H₂O ⇄ H₃O⁺ + C₂H₃O₂⁻</em>
f. As the acetic acid (HC₂H₃O₂) is a week acid, the dissociation in C₂H₃O₂⁻ is not complete, that means that <em>ka<<1</em>
g. The ka for this reaction is 1,8x10⁻⁵, that means that there are more <em>weak acid molecules</em> (HC₂H₃O₂) than conjugate base ions. Also, the <em>water molecules </em>will be in higher proportion than hydronium ions.
I hope it helps!
