Answer:
ai) Rate law, ![Rate = k [CH_3 Cl] [Cl_2]^{0.5}](https://tex.z-dn.net/?f=Rate%20%3D%20k%20%5BCH_3%20Cl%5D%20%5BCl_2%5D%5E%7B0.5%7D)
aii) Rate constant, k = 1.25
b) Overall order of reaction = 1.5
Explanation:
Equation of Reaction:

If
, the rate of backward reaction is given by:
![Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}](https://tex.z-dn.net/?f=Rate%20%3D%20k%20%5BA%5D%5E%7Ba%7D%20%5BB%5D%5E%7Bb%7D%5C%5Ck%20%3D%20%5Cfrac%7BRate%7D%7B%20%5BA%5D%5E%7Ba%7D%20%5BB%5D%5E%7Bb%7D%7D%5C%5Ck%20%3D%20%5Cfrac%7BRate%7D%7B%20%5BCH_3%20Cl%5D%5E%7Ba%7D%20%5BCl_2%5D%5E%7Bb%7D%7D)
k is constant for all the stages
Using the information provided in lines 1 and 2 of the table:
![0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1](https://tex.z-dn.net/?f=0.014%20%2F%20%5B0.05%5D%5Ea%20%5B0.05%5D%5Eb%20%3D%2000.029%2F%20%5B0.100%5D%5Ea%20%5B0.05%5D%5Eb%5C%5C0.014%20%2F%20%5B0.05%5D%5Ea%20%5B0.05%5D%5Eb%20%3D%2000.029%2F%20%5B2%2A0.05%5D%5Ea%20%5B0.05%5D%5Eb%5C%5C0.014%20%2F%20%3D%200.029%2F%202%5Ea%5C%5C2%5Ea%20%3D%202.07%5C%5Ca%20%3D%201)
Using the information provided in lines 3 and 4 of the table and insering the value of a:
![0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\](https://tex.z-dn.net/?f=0.041%20%2F%20%5B0.100%5D%5Ea%20%5B0.100%5D%5Eb%20%3D%200.115%20%2F%20%5B0.200%5D%5Ea%20%5B0.200%5D%5Eb%5C%5C0.041%20%2F%20%5B0.100%5D%5Ea%20%5B0.100%5D%5Eb%20%3D%200.115%20%2F%20%5B2%20%2A%200.100%5D%5Ea%20%5B2%20%2A%200.100%5D%5Eb%5C%5C)
![0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5](https://tex.z-dn.net/?f=0.041%20%3D%200.115%20%2F%20%5B2%20%5D%5Ea%20%5B2%5D%5Eb%5C%5C%20%5C%5B%5B2%20%5D%5Ea%20%5B2%5D%5Eb%20%3D%200.115%2F0.041%5C%5C%20%5C%5B%5B2%20%5D%5Ea%20%5B2%5D%5Eb%20%3D%202.80%5C%5C%5C%5B%5B2%20%5D%5E1%20%5B2%5D%5Eb%20%3D%202.80%5C%5C%5C%5B%5B2%5D%5Eb%20%3D%201.40%5C%5Cb%20%3D%20%5Cfrac%7Bln%201.4%7D%7Bln%202%7D%20%5C%5Cb%20%3D%200.5)
The rate law is: ![Rate = k [CH_3 Cl] [Cl_2]^{0.5}](https://tex.z-dn.net/?f=Rate%20%3D%20k%20%5BCH_3%20Cl%5D%20%5BCl_2%5D%5E%7B0.5%7D)
The rate constant
then becomes:
![k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25](https://tex.z-dn.net/?f=k%20%3D%200.014%20%2F%20%28%20%5B0.050%5D%20%5B0.050%5D%5E%280.5%29%20%29%5C%5Ck%20%3D%201.25)
b) Overall order of reaction = a + b
Overall order of reaction = 1 + 0.5
Overall order of reaction = 1.5
The first one is correct cause first of all it's atomic number is 7 and second of all it's atomic mass is about 15
A hydrate is a chemical that has water molecules loosely bonded to it. The water molecules are not ... You will be using the hydrate CuSO4 . ?H2O. Sample Calculation-. An empty crucible has a mass of 12.770 grams.
Answer:
Lithium selenide Lithium selenide (Li2Se) 12136-60-6 Dilithium selenide lithium selenidolithium
Molecular Weight 92.9 g/mol
Dates Modify 2020-11-15 Create 2005-08-08
The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C