Answer:
Kp = 1.41 x 10⁻⁶
Explanation:
We have the chemical equation:
2 A(g) + 3 B(g)⇌ C(g)
In which A and B are the reactants and C is the product. We calculate first the change in the number of moles of gas (Δn or dn):
dn= (sum moles products - sum moles reactants)
= (moles C - (moles A + moles B))
= (1 - (2+3))
= 1 - 5
= -4
We have also the following data:
Kc = 63.2
T= 81∘C + 273 = 354 K
R = 0.082 L.atm/K.mol (it is a constant)
Thus, we introduce the data in the mathematical expression for the relation between Kp and Kc:
= (0.082 L.atm/K.mol x 354 K)⁻⁴ = 1.41 x 10⁻⁶
Answer:
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Explanation:
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Chemical equations are to be balanced to be able to follow the law of conservation of mass where it says that mass cannot be created or destroyed. Reactions should be that the mass of the reactants is equal to the mass of the products.
Answer : The correct option is, (b) +0.799 V
Solution :
The values of standard reduction electrode potential of the cell are:
![E^0_{[H^{+}/H_2]}=+0.00V](https://tex.z-dn.net/?f=E%5E0_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D%3D%2B0.00V)
![E^0_{[Ag^{+}/Ag]}=+0.799V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.799V)
From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) : 
Reaction at cathode (reduction) : 
The balanced cell reaction will be,
Now we have to calculate the standard electrode potential of the cell.
![E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D)
Therefore, the standard cell potential will be +0.799 V
