Answer:
Hi! Think of it like this:
You and your family are going to go on a vaction for the holidays, to south america or somewhere else. You´re near the equator so it will be warm. If you were to move there, you might pull out the swimsiutes because most of the time, its useally warm and sunny. They dont really get cold weather. They will have a nice and warm christmas.
The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
![rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
At a certain concentration of ![H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=H_2%20and%20%5Btex%5DI_2%2C%20the%20initial%20rate%20of%20reaction%20is%200.120%20M%2Fs.%20What%20would%20the%20initial%20rate%20of%20the%20reaction%20be%20if%20the%20concentration%20of%20%5Btex%5DH_2%20were%20halved.%3C%2Fp%3E%3Cp%3E%3Cstrong%3EAnswer%20%3A%20The%20initial%20rate%20of%20the%20reaction%20will%20be%2C%200.03%20M%2Fs%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%20%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3ERate%20law%20expression%20for%20the%20reaction%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%5Btex%5Drate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
As we are given that:
Initial rate = 0.120 M/s
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Dividing 2 by 1, we get:
![\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B0.120%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D%7D%7Bk%5BH_2%5D%5E2%5BNH_3%5D%7D)


Therefore, the initial rate of the reaction will be, 0.03 M/s