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[I_2]^2](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BI_2%5D%5E2)
At a certain concentration of 
and 
, the initial rate of reaction is 4.0 × 10⁴ M/s. What would the initial rate of the reaction be if the concentration of
Answer : The initial rate of the reaction will be, 
Explanation :
Rate law expression for the reaction:
As we are given that:
Initial rate = 4.0 × 10⁴ M/s
Expression for rate law for first observation:
![4.0\times 10^4=k[H_2]^2[I_2]^2](https://tex.z-dn.net/?f=4.0%5Ctimes%2010%5E4%3Dk%5BH_2%5D%5E2%5BI_2%5D%5E2)
....(1)
Expression for rate law for second observation:
![R=k(\frac{[H_2]}{2})^2[I_2]^2](https://tex.z-dn.net/?f=R%3Dk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BI_2%5D%5E2)
....(2)
Dividing 2 by 1, we get:
![\frac{R}{4.0\times 10^4}=\frac{k(\frac{[H_2]}{2})^2[I_2]^2}{k[H_2]^2[I_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B4.0%5Ctimes%2010%5E4%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BI_2%5D%5E2%7D%7Bk%5BH_2%5D%5E2%5BI_2%5D%5E2%7D)
Therefore, the initial rate of the reaction will be,
Answer:3.5 moles * 6.02 X 10^23 particles/mole * 2 H atoms/particle of H2 = 42.14 X 10^^23 atoms
Explanation:
In exothermic reactions, there is a release heat and the replacement of weak bonds with stronger ones.
Answer:
Hydrogen concentration = 7.94×10^-3 M
Explanation:
from potenz Hydrogen ( pH ) definition
pH = -log[H+]
2.1 = -log[H+]
2.1/-log = -log[H+]/-log
10^-2.1 = [H+]
[H+] = 7.94×10^-3M
Answer:
6.28
Explanation:
Let's consider the following reaction at equilibrium.
CH₄(g) + H₂O(g) ⇌ 3 H₂(g) + CO(g)
The concentration equilibrium constant (Kc) is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients.
Kc = [H₂]³ × [CO] / [CH₄] × [H₂O]
Kc = 1.15³ × 0.126 / 0.126 × 0.242
Kc = 6.28
