Question

for the reaction 3a + 4b → 2c + d, what is the magnitude of the rate of change for b when d is increasing at 2.4 m/min? answer: m/min

Explanation:

Step1: Determine rate - relationship

For the reaction \(3A + 4B
ightarrow2C + D\), the rate of the reaction can be expressed in terms of the reactants as \(rate=-\frac{1}{3}\frac{\Delta[A]}{\Delta t}=-\frac{1}{4}\frac{\Delta[B]}{\Delta t}\).

Step2: Solve for \(\frac{\Delta[B]}{\Delta t}\)

We know that \(\frac{\Delta[A]}{\Delta t} = 2.4\ M/min\). From the rate - relationship \(-\frac{1}{3}\frac{\Delta[A]}{\Delta t}=-\frac{1}{4}\frac{\Delta[B]}{\Delta t}\), we can cross - multiply to get \(\frac{\Delta[B]}{\Delta t}=\frac{4}{3}\times\frac{\Delta[A]}{\Delta t}\).

Step3: Substitute the value of \(\frac{\Delta[A]}{\Delta t}\)

Substitute \(\frac{\Delta[A]}{\Delta t} = 2.4\ M/min\) into the equation \(\frac{\Delta[B]}{\Delta t}=\frac{4}{3}\times2.4\ M/min\). Then \(\frac{\Delta[B]}{\Delta t}=3.2\ M/min\).

Answer:

\(3.2\ M/min\)