Question

1. for the decomposition reaction 2ab → a₂ + b₂, the concentration of a₂ after 4 min = 0.52 mol/l and after 7.5 min its concentration is 0.68 mol/l. what is the average rate of change in the concentration of ab over this period of time? ( /3)

Explanation:

Step1: Recall the reaction stoichiometry

The reaction is \(2AB
ightarrow A_2 + B_2\). From the stoichiometry, for every 1 mole of \(A_2\) formed, 2 moles of \(AB\) are consumed. So the change in concentration of \(AB\) is related to the change in concentration of \(A_2\) by \(\Delta[AB]= - 2\Delta[A_2]\) (negative because \(AB\) is a reactant, its concentration decreases).

Step2: Calculate \(\Delta[A_2]\)

\(\Delta[A_2]=[A_2]_{final}-[A_2]_{initial}\). Here, \(t_1 = 4\space min\), \([A_2]_1 = 0.52\space mol/L\); \(t_2=7.5\space min\), \([A_2]_2 = 0.68\space mol/L\). So \(\Delta[A_2]=0.68 - 0.52=0.16\space mol/L\).

Step3: Calculate \(\Delta[AB]\)

Using the stoichiometric ratio, \(\Delta[AB]=- 2\times\Delta[A_2]\). Substituting \(\Delta[A_2] = 0.16\space mol/L\), we get \(\Delta[AB]=-2\times0.16=- 0.32\space mol/L\) (negative because \(AB\) is consumed).

Step4: Calculate the time interval \(\Delta t\)

\(\Delta t=t_2 - t_1=7.5 - 4 = 3.5\space min\).

Step5: Calculate the average rate of change of \(AB\)

The average rate of change of a substance is given by \(\text{Rate}=\frac{\Delta[AB]}{\Delta t}\). Substituting \(\Delta[AB]=- 0.32\space mol/L\) and \(\Delta t = 3.5\space min\), we get \(\text{Rate}=\frac{- 0.32}{3.5}\space mol/(L\cdot min)\). But we take the magnitude for the rate of change (since rate of change of reactant concentration is negative, but we can also consider the absolute value for the rate of consumption). Wait, actually, the average rate of change of \(AB\) is \(\frac{\Delta[AB]}{\Delta t}\). Let's compute it: \(\frac{- 0.32}{3.5}\approx - 0.0914\space mol/(L\cdot min)\). But let's check again. Wait, the change in \(A_2\) is \(0.68 - 0.52 = 0.16\space mol/L\) over \(\Delta t=7.5 - 4=3.5\space min\). The rate of formation of \(A_2\) is \(\frac{\Delta[A_2]}{\Delta t}=\frac{0.16}{3.5}\space mol/(L\cdot min)\). Then, from the reaction, the rate of consumption of \(AB\) is \(2\times\) rate of formation of \(A_2\) (because 2 moles of \(AB\) produce 1 mole of \(A_2\)). So rate of change of \(AB\) is \(-2\times\frac{\Delta[A_2]}{\Delta t}\) (negative because it's a reactant). So \(\text{Rate of change of }AB=\frac{\Delta[AB]}{\Delta t}=-2\times\frac{0.68 - 0.52}{7.5 - 4}\).

Step6: Compute the value

First, calculate the numerator for \(A_2\) change: \(0.68 - 0.52 = 0.16\space mol/L\). Time change: \(7.5 - 4 = 3.5\space min\). Then, rate of \(A_2\) formation: \(\frac{0.16}{3.5}\space mol/(L\cdot min)\). Then, rate of \(AB\) consumption: \(2\times\frac{0.16}{3.5}=\frac{0.32}{3.5}\approx0.0914\space mol/(L\cdot min)\), but since \(AB\) is a reactant, its rate of change is negative, so \(\frac{\Delta[AB]}{\Delta t}=- \frac{0.32}{3.5}\approx - 0.091\space mol/(L\cdot min)\) (or we can say the average rate of change in concentration of \(AB\) is \(- 0.091\space mol/(L\cdot min)\), and the magnitude of the rate of consumption is \(0.091\space mol/(L\cdot min)\)). Wait, let's do the calculation precisely:

\(\Delta t=7.5 - 4 = 3.5\space min\)

\(\Delta[A_2]=0.68 - 0.52 = 0.16\space mol/L\)

From the reaction, \(2\space mol\) of \(AB\) produces \(1\space mol\) of \(A_2\), so \(\Delta[AB]=- 2\times\Delta[A_2]\) (the negative sign indicates decrease in concentration of \(AB\))

So \(\Delta[AB]=- 2\times0.16=- 0.32\space mol/L\) over \(\Delta t = 3.5\space min\)

Then the average rate of change of \(AB\) is \(\frac{\Delta[AB]}{\Delta t}=\frac{- 0.32}{3.5}\approx - 0.091\space mol/(L\cdot min)\) (or approximately \(- 0.091\space M/min\), where \(M = mol/L\))

Answer:

The average rate of change in the concentration of \(AB\) is \(\boldsymbol{- \frac{16}{175}\space mol/(L\cdot min)\approx - 0.091\space mol/(L\cdot min)}\) (or if we consider the magnitude of the rate of consumption, it's \(\approx0.091\space mol/(L\cdot min)\), but the rate of change (which includes the sign for reactant) is approximately \(- 0.091\space mol/(L\cdot min)\))