For the reaction 4 A + 5 B → 2 C + 3 D, what is the magnitude of the rate of change for [D] when [A] is decreasing at 0.930 M/s?

Answer 1

The given balanced chemical equation is:

4 A + 5 B → 2 C + 3 D

We are given that [A] is decreasing at a rate of 0.930 M/s. Let's assume that the rate of the reaction is

expressed in terms of the rate of change of [D], i.e., d[D]/dt.

Using the stoichiometry of the reaction, we can write:

(d[D]/dt) = (3/4) × (Rate of disappearance of A)

where (d[D]/dt) is the rate of change of concentration of D, and (Rate of disappearance of A) is the rate at

which A is consumed in the reaction.

We are given that (Rate of disappearance of A) = -0.930 M/s (negative sign indicates that [A] is decreasing).

Substituting this value in the above equation, we get:

(d[D]/dt) = (3/4) × (-0.930 M/s) = -0.698 M/s

Therefore, the magnitude of the rate of change for [D] is 0.698 M/s, and it is negative, indicating that the concentration of D is decreasing with time.