for the reaction 2a + 3b → 4d + 5e, how is the rate of reaction expressed in terms of the rate of…

Explanation:

Step1: Recall reaction - rate formula

For a general reaction (aA + bB\rightarrow cC + dD), the rate of the reaction (r=-\frac{1}{a}\frac{\Delta[A]}{\Delta t}=-\frac{1}{b}\frac{\Delta[B]}{\Delta t}=\frac{1}{c}\frac{\Delta[C]}{\Delta t}=\frac{1}{d}\frac{\Delta[D]}{\Delta t}). For the reaction (2A + 3B\rightarrow4D+5E), the rate of the reaction (r =-\frac{1}{2}\frac{\Delta[A]}{\Delta t}=-\frac{1}{3}\frac{\Delta[B]}{\Delta t}=\frac{1}{4}\frac{\Delta[D]}{\Delta t}=\frac{1}{5}\frac{\Delta[E]}{\Delta t}).

Step2: Express rate in terms of B

We want to express the rate of the reaction in terms of the rate of disappearance of B. The rate of disappearance of B is (-\frac{\Delta[B]}{\Delta t}). The rate of the reaction (r =-\frac{1}{3}\frac{\Delta[B]}{\Delta t}), so (-\frac{\Delta[B]}{\Delta t}=3r). The rate of the reaction can be written as (r =-\frac{\Delta[B]}{3\Delta t}).

Answer:

A. (\frac{\Delta[B]}{\Delta t}) is incorrect because the rate of disappearance of B has a negative sign. B. (-\frac{\Delta[B]}{\Delta t}) represents the rate of disappearance of B but not the rate of the reaction in the correct form. C. (\frac{1}{3}\frac{\Delta[B]}{\Delta t}) is incorrect as it has the wrong sign for the reactant - B. D. (-\frac{1}{3}\frac{\Delta[B]}{\Delta t}) is the correct expression for the rate of the reaction in terms of the rate of disappearance of B. E. (\frac{3}{\Delta t}\Delta[B]) is incorrect. So the answer is D. (-\frac{1}{3}\frac{\Delta[B]}{\Delta t})