Consider the following reaction was found to be the 0th order to A and 2nd order to B.
2A(cr) + 3B(aq) -->A2B3(cr)
Write a rate law equation for the reaction rate of A with B.
rate = k*[A]o[B]2
OR you may omit the A terms since A raised to the zero power is 1. The "trick" to the problem, I suspect, is that many will write it as (A)^2*(B)^3.
Ok, thanks.
Couple follow ups:
Q1: If [A] is doubled how does the reaction rate change?
A: I put no change because 0th order cancels it out.
Q2: If [B] is doubled how does the reaction rate change?
A: I put it increases by a factor of 4 because of the squared term.
Q3: How would increase pressure effect the reaction rate?
A:???
Now that I look at Q3 I am thinking it is going to increase it, but not sure by what factor?
The answer to A is correct since it's zero order with respect to A. The answer to B is correct since 2^2 = 4.
Look at the equation. A is Cr which I assume means crystalline and that means a solid. B is aq which I assume means a liquid and A2B3 is cr which again I assume is a solid. Pressure will have an effect if reactants or products are gases; however, pressure has no effect on solids and essentially no effect on liquids. So increased pressure, I think, will have no effect on the rate of this reaction. .
To write the rate law equation for the reaction rate of A with B, we need to determine the order of the reaction with respect to A and B.
It is given that the reaction is 0th order with respect to A, which means the concentration of A does not affect the rate of the reaction. Therefore, A does not appear in the rate law equation.
On the other hand, it is given that the reaction is 2nd order with respect to B, which means the rate of the reaction is directly proportional to the square of the concentration of B.
So, the rate law equation for the reaction rate of A with B is:
Rate = k[B]^2
Here, [B] represents the concentration of B, and k is the rate constant for the reaction.