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?
for this one do i just plug in k A^2B^0 = rate?
yes
To calculate the rate of the reaction, we need to use the rate law equation:
Rate = k[A]^2[B]^0
Given that the reaction is second order in A and zero order in B, the rate law equation can be simplified as follows:
Rate = k[A]^2
Plugging in the values, we get:
Rate = 1.10×10^−2 M^−1 min^−1 * (0.124 M)^2
Calculating the value:
Rate = 1.10×10^−2 M^−1 min^−1 * (0.124 M)^2
Rate = 1.10×10^−2 * 0.015376 M^−1 min^−1
Rate ≈ 1.6928 × 10^−4 M/min
Therefore, the rate of the reaction when [A] = 0.124 M and [B] = 3.89 M is approximately 1.6928 × 10^−4 M/min.
To find the rate of the reaction, we can use the rate equation for a second-order reaction:
rate = k * [A]^2 * [B]^0
Since the order of B is zero, its concentration doesn't affect the rate of the reaction.
Given:
[A] = 0.124 M
[B] = 3.89 M
k = 1.10×10^(-2) M^(-1) min^(-1)
Substituting the values into the rate equation, we have:
rate = k * [A]^2 * [B]^0
= (1.10×10^(-2) M^(-1) min^(-1)) * (0.124 M)^2 * (3.89 M)^0
= (1.10×10^(-2) M^(-1) min^(-1)) * (0.124 M)^2
Calculating this value gives us the rate of the reaction.