Answer:
The rate equation for this reaction:
![R=k[NH_3]^0](https://tex.z-dn.net/?f=R%3Dk%5BNH_3%5D%5E0)
Explanation:
Decomposition of ammonia:
Rate law of the can be written as;
![R=k[NH_3]^x](https://tex.z-dn.net/?f=R%3Dk%5BNH_3%5D%5Ex)
1) Rate of the reaction , when ![[NH_3]=2.0\times 10^{-3} M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D2.0%5Ctimes%2010%5E%7B-3%7D%20M)
![1.5\times 10^{-6} M/s=k[2.0\times 10^{-3} M]^x](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-6%7D%20M%2Fs%3Dk%5B2.0%5Ctimes%2010%5E%7B-3%7D%20M%5D%5Ex)
..[1]
2) Rate of the reaction , when ![[NH_3]=4.0\times 10^{-3} M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D4.0%5Ctimes%2010%5E%7B-3%7D%20M)
![1.5\times 10^{-6} M/s=k[4.0\times 10^{-3} M]^x](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-6%7D%20M%2Fs%3Dk%5B4.0%5Ctimes%2010%5E%7B-3%7D%20M%5D%5Ex)
..[2]
[1] ÷ [2]
![\frac{1.5\times 10^{-6}M/s}{1.5\times 10^{-6}M/s}=\frac{k[2.0\times 10^{-3}M]^x}{k[4.0\times 10^{-3}M]^x}](https://tex.z-dn.net/?f=%5Cfrac%7B1.5%5Ctimes%2010%5E%7B-6%7DM%2Fs%7D%7B1.5%5Ctimes%2010%5E%7B-6%7DM%2Fs%7D%3D%5Cfrac%7Bk%5B2.0%5Ctimes%2010%5E%7B-3%7DM%5D%5Ex%7D%7Bk%5B4.0%5Ctimes%2010%5E%7B-3%7DM%5D%5Ex%7D)
On solving for x , we get ;
x = 0
The rate equation for this reaction:
Answer:
Phenol.
Explanation:
Hello there!
In this case, for this question about the strength of the resulting conjugate base, it is possible for us to know that the higher the Ka (stronger aid) the weaker the conjugate base. In such a way, since the phenol has the smallest Ka (weakest acid) we infer that it has the strongest conjugate base.
In addition to the aforementioned, the reason why phenol was the answer is because the Kb of its conjugate base is the greatest due to the Kb=Kw/Ka and thus, the smaller the Ka the bigger Kb.
Regards!
