HNO₃ + H₂S → S + NO + H₂<span>O
Assign Oxidation Number:
L.H.S R.H.S
N in HNO</span>₃ = +5 +2 = N in NO
S in H₂S = -2 0 = S in S
Write Half cell Reactions:
Reduction Reaction:
3e⁻ + HNO₃ → NO -------(1)
Oxidation Reaction:
H₂S → S + 2e⁻ -------(2)
Multiply eq. 1 with 2 and eq. 2 with 3 to balance electrons.
6e⁻ + 2 HNO₃ → 2 NO
3 H₂S → 3 S + 6e⁻
Cancel e⁻s,
______________________________
2 HNO₃ + 3 H₂S → 2 NO + 3 S + H₂O
Balance Oxygen Atoms by multiplying H₂O with 4, Hydrogen atoms will automatically get balance.
2 HNO₃ + 3 H₂S → 2 NO + 3 S + 4H₂O
Answer:
The new equilibrium total pressure will be increased to one-half to initial total pressure.
Explanation:
From the information given :
The equation of the reaction can be represented as;
From above equation:
2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide
So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.
So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.
Let the total pressure at the initial equilibrium be 
and the total pressure at the final equilibrium be 
According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Thus;
P ∝ 1/V
P = K/V
PV = K
where K = constant
So;
PV = constant
Hence;
From the foregoing; since the volume is decreased to one- half to initial Volume; then ,
also;
Thus ;
Dividing both sides by 
From ;
Thus; The new equilibrium total pressure will be increased to one-half to initial total pressure.
To convert from moles to grams you divide by the molar mass of the element. To convert from grams to moles you X by the molar mass element
x= the coefficients in front of the substance in the balanced chemical equation
[H+]= the concentration of hydrogen ions
[A-]= the concentration of the other ion that broke off from the H+
[HA]= the un-disassociated acid concentration
The higher the Ka value, the greater amount of disassociation of the reactants into products. As for acids, they will break down to form H+ ions. The more the H+ ions, the stronger acidity of the solution. Thus since A has the highest Ka value, that represents the strongest acid.
You can determine the Ka value from a number of ways. If equilibrium concentrations are given of a certain acid solution, you can find the proportion of the concentration of ions to the concentration of the remaining HA molecules, using the equation above. Also, pH and KpH can be used in a number of ways. This gets more complicated and depends on the situation, and requires more advanced equations.
Hope this helped a little, its obviously not my best work
