The reaction A + B to C + D is second order in A and zero order in B. The value of k is 1.10×10−2 M^{-1}\;min^{-1}.
What is the rate of this reaction when A}]=0.124 M and B}]=3.89 M?
To find the rate of the reaction, we need to use the rate equation:
Rate = k[A]^n[B]^m
According to the given information, the reaction is second order in A (n = 2) and zero order in B (m = 0). This means that the concentration of A is squared in the rate equation, and B has no effect on the rate.
The value of k is given as 1.10×10−2 M^{-1} min^{-1}.
Now we can substitute the given concentrations into the rate equation:
Rate = k[A]^2[B]^0
Since [B]^0 is equal to 1, we can simplify it further:
Rate = k[A]^2
Substituting the given concentrations:
Rate = (1.10×10^{-2} M^{-1} min^{-1}) x (0.124 M)^2
Calculating the value:
Rate = (1.10×10^{-2} M^{-1} min^{-1}) x (0.124 M)^2
= 1.10×10^{-2} M^{-1} min^{-1} x 0.015376 M^2
= 1.6876×10^{-4} M min^{-1}
Therefore, the rate of the reaction when [A] = 0.124 M and [B] = 3.89 M is 1.6876×10^{-4} M min^{-1}.