the rate of the following reaction is 0.72 M/s. what is the relative rate of change of each species in the reaction?
A+4B -> 2C
rate of change is dA/dT = (1/4)dB/dT = (1/2)dC/dT = 0.72
Solve for dA/dT, dB/dT, dC/dT
In general, rate is equal to -A=-1/4(B)=1/2(C)
Assuming 0.72 M/s is the appearance of C
then A's disappearance is = to 1/2(0.72 M/s) and B's is = to 4/2(0.72 M/s)
A and B will be negative.
Didn't know that you were answering this.
^i was told that this is wrong...
Not sure who told you that but I am pretty sure that Dr.bob222 and I were both correct.
To find the relative rate of change of each species in a chemical reaction, we can use stoichiometry and the rate expression.
The balanced equation for the reaction A + 4B -> 2C tells us that for every one molecule of A reacted, four molecules of B are also consumed, and two molecules of C are produced.
The rate expression for this reaction is given as:
Rate = -1/4 * Δ[A]/Δt = Δ[B]/Δt = 1/2 * Δ[C]/Δt
Here, Δ[A], Δ[B], and Δ[C] represent the change in concentration of A, B, and C over a specific time interval Δt.
Given that the rate of the reaction is 0.72 M/s, we can now calculate the relative rate of change for each species:
For species A:
Relative rate of change = -1/4 * Δ[A]/Δt = -1/4 * 0.72 M/s = -0.18 M/s
For species B:
Relative rate of change = Δ[B]/Δt = 0.72 M/s
For species C:
Relative rate of change = 1/2 * Δ[C]/Δt = 1/2 * 0.72 M/s = 0.36 M/s
Therefore, the relative rate of change of species A is -0.18 M/s, the relative rate of change of species B is 0.72 M/s, and the relative rate of change of species C is 0.36 M/s.