Respuesta :
The given question is incomplete. The complete question is :
For the net reaction: [tex]2AB+2C\rightarrow A_2+2BC[/tex], the following slow first steps have been proposed.
A. [tex]2C+AB\rightarrow AC+BC[/tex]
B. [tex]2AB\rightarrow A_2+2B[/tex]
C. [tex]C+AB\rightarrow BC+A[/tex]
D. [tex]AB \rightarrow A + B[/tex]
Determine the units of the rate constant for all four reactions listed in the problem above, and enter the correct choices from the list below. Enter 4 letters in order (e.g. ABCD or CBED)
a. [tex]moleL^{-1 }sec^{-1}[/tex]
b. [tex]mole^{-1}Lsec^{-1}[/tex]
c. [tex]mole^2L^{-2}sec^{-1}[/tex]
d. [tex]mole^{-2}L^2sec^{-1}[/tex]
e. None of the above.
Answer: A. [tex]2C+AB\rightarrow AC+BC[/tex] : [tex]mol^{-2}L^2s^{-1}[/tex]
B. [tex]2AB\rightarrow A_2+2B[/tex] : [tex]k=mol^{-1}Ls^{-1}[/tex]
C. [tex]C+AB\rightarrow BC+A[/tex] : [tex]k=mol^{-1}Ls^{-1}[/tex]
D. [tex]AB \rightarrow A + B[/tex]: [tex]k=s^{-1}[/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
1. [tex]2C+AB\rightarrow AC+BC[/tex]
[tex]Rate=k[C]^2[AB}^1[/tex]
[tex]molL^{-1}s^{-1}=k[molL^{-1}]^2[molL^{-1}}^1[/tex]
[tex]k=mol^{-2}L^2s^{-1}[/tex]
2. [tex]2AB\rightarrow A_2+2B[/tex]
[tex]Rate=k[AB]^2[/tex]
[tex]molL^{-1}s^{-1}=k[molL^{-1}]^2[/tex]
[tex]k=mol^{-1}Ls^{-1}[/tex]
3. [tex]C+AB\rightarrow BC+A[/tex]
[tex]Rate=k[C}^1[AB]^1[/tex]
[tex]molL^{-1}s^{-1}=k[molL^{-1}]^1[molL^{-1}]^1[/tex]
[tex]k=mol^{-1}Ls^{-1}[/tex]
4. [tex]AB \rightarrow A + B[/tex]
[tex]Rate=k[AB]^1[/tex]
[tex]molL^{-1}s^{-1}=k[molL^{-1}]^1[/tex]
[tex]k=s^{-1}[/tex]