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

Explanation:

Step1: Recall rate - reaction formula

For a general reaction (aA + bB\rightarrow cC + dD), the rate of the reaction is given by (rate=-\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. From (r =-\frac{1}{3}\frac{\Delta[B]}{\Delta t}), we can solve for (r). The rate of the reaction (r=-\frac{1}{3}\frac{\Delta[B]}{\Delta t}), which can be rewritten as (\frac{1}{3}\left(-\frac{\Delta[B]}{\Delta t}\right)). The negative sign indicates the decrease in the concentration of B over time ((\Delta[B]<0) as B is consumed). The rate of disappearance of B is (-\frac{\Delta[B]}{\Delta t}), and the rate of the reaction (r = \frac{1}{3}\left(-\frac{\Delta[B]}{\Delta t}\right)).

Answer:

C. (\frac{1}{3}\left(-\frac{\Delta[B]}{\Delta t}\right))