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 a reaction, you need to identify the order of each reactant in the reaction equation. The order of a reactant represents how its concentration affects the rate of the reaction.
Given the reaction equation:
4A + 2B --> 6C
We can determine the order of each reactant by comparing the changes in their concentrations in the reaction equation.
For A, we see that its concentration decreases by a factor of 2 (from 4A to 2A), while the rate of the reaction increases by a factor of 3 (from 3C to 6C). This indicates that A has a second-order relationship with the rate of the reaction.
For B, we see that its concentration decreases by a factor of 1 (from 2B to 1B), while the rate of the reaction increases by a factor of 3 (from 3C to 6C). This suggests that B has a first-order relationship with the rate of the reaction.
Based on this information, we can write the rate law equation for the reaction as:
Rate = k[A]^2[B]
Where:
- Rate represents the rate of the reaction
- k is the rate constant, which depends on temperature and other factors
- [A] represents the concentration of A
- [B] represents the concentration of B
Therefore, the rate law equation for the given reaction is:
Rate = k[A]^2[B]