Given :
2NOBr(g) - -> 2NO(g) + Br2(g)
Initial pressure of NOBr , 1 atm .
At equilibrium, the partial pressure of NOBr is 0.82 atm.
To Find :
The equilibrium constant for the reaction .
Solution :
2NOBr(g) - -> 2NO(g) + Br2(g)
t=0 s 1 atm 0 0
1( 1-2x) 2x x
So ,
At equilibrium :
![K_{eq}=\dfrac{[NO]^2[br_2]}{[NOBr]^2}\\\\K_{eq}=\dfrac{0.18^2\times 0.9}{0.82^2}\\\\K_{eq}=0.043\ atm](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cdfrac%7B%5BNO%5D%5E2%5Bbr_2%5D%7D%7B%5BNOBr%5D%5E2%7D%5C%5C%5C%5CK_%7Beq%7D%3D%5Cdfrac%7B0.18%5E2%5Ctimes%200.9%7D%7B0.82%5E2%7D%5C%5C%5C%5CK_%7Beq%7D%3D0.043%5C%20atm)
Hence , this is the required solution .
Answer:
The value of the carbon bond angles are 109.5 °
Explanation:
CH3CH2CH2OH = propanol . This is an alcohol.
All bonds here are single bonds.
Single bonds are sp³- hybdridization type. To be sp3 hybridized, it has an s orbital and three p orbitals : sp³. This refers to the mixing character of one 2s-orbital and three 2p-orbitals. This will create four hybrid orbitals with similar characteristics.
Sp3- types have angles of 109.5 ° between the carbon - atoms.
This means that the value of the carbon bond angles are 109.5 °
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
