The following reaction is second order in A and first order in B.
2A+1B --> 3C
what is the rate law equation for the reaction below?
(Assume that A, B and C are the same compounds for each reaction.)
4A + 2B --> 6C
To determine the rate law equation for the given reaction, we need to analyze the stoichiometry of the reaction and the order of each reactant.
From the given reaction equation:
2A + 1B --> 3C
We can see that the coefficient of A is 2, implying that A is involved in a second-order reaction. Similarly, the coefficient of B is 1, indicating that B is involved in a first-order reaction.
Therefore, we can write the rate law equation for the reaction as follows:
Rate = k[A]^2[B]^1
Where:
- Rate represents the rate of the reaction
- k is the rate constant, which is specific to the reaction at a given temperature
- [A] represents the concentration of A
- [B] represents the concentration of B
Now, let's determine the rate law equation for the second reaction:
4A + 2B --> 6C
By inspecting the coefficients, we can see that the coefficient of A is 4, indicating a second-order reaction for A. Likewise, the coefficient of B is 2, suggesting a first-order reaction for B.
Hence, the rate law equation for the second reaction can be written as:
Rate = k'[A]^4[B]^2
Where:
- Rate represents the rate of the reaction
- k' is the rate constant specific to the second reaction at a given temperature
- [A] represents the concentration of A
- [B] represents the concentration of B
Using this method, you can determine the rate law equation for any given reaction by examining the stoichiometric coefficients of the reactants.