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:
![0.120=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=0.120%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
....(1)
Expression for rate law for second observation:
![R=k(\frac{[H_2]}{2})^2[NH_3]](https://tex.z-dn.net/?f=R%3Dk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D)
....(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
Answer:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
2NO(g) + O₂(g) → 2 NO₂
Explanation:
First of all, we need to consider the reaction for production of ammonia. In this reaction we have as reactants, nitrogen and hydroge.
3H₂ (g) + N₂(g) → 2NH₃ (g)
Afterwards, ammonia reacts to oxygen, to produce NO and H₂O
The equation for the process will be:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
Then, we take the nitric oxide to make it react, to produce NO₂, in order to produce nitric acid, for the final reaction:
2NO(g) + O₂(g) → 2 NO₂
3NO₂(g) + H₂O(g) → 2 HNO₃ (g) + NO(g)
Answer:
The average kinetic energy of the system has increased as a result of the temperature increasing.
Explanation:
Assuming this is a gas based on the framing.
The molecules of a gas span a distribution of speeds, and the average kinetic energy of the molecules is directly proportional to the absolute temperature of the sample. KEavg is proportional to T.
This can be further studied until the Kinetic-Molecular Theory.
