In a reaction \(2A \longrightarrow B\), the concentration of \(A\) decreases from \(0.64\, \ M\) to \(0.22\, \ M\) is \(5\) minutes. What is the rate of reaction (in \(Ms^{-1}\)) during this interval?
Answer & explanation
Correct answer: option 2
The correct answer is option 2. \(14 × 10^{-4}\, \ Ms^{-1}\).
We can solve this problem by finding the rate of disappearance of reactant \(A\), which is directly proportional to the rate of the reaction.
\(\Delta [A] = [A]_{initial}\, \ -\, \ [A]_{final} = 0.64 M\, \ -\, \ 0.22 M = 0.42 M\)
\(\Delta t = 5 \text{ minutes} = 5 \text{ minutes} × (60 \text{seconds/minute}) = 300 \text{ seconds}\)
The rate of disappearance of A is given by:
\(\text{ Rate of A = } \frac{-\Delta [A]}{\Delta t}\)
\(\text{ Rate of A = } \frac{-0.42 M}{300 s} = -1.4 × 10^{-3}\, \ Ms^{-1}\)
However, the rate is defined as a positive value, so we take the absolute value:
\(\text{ Rate of A = } 1.4 × 10^{-3}\, \ Ms^{-1} = 14 × 10^{-4}\, \ Ms^{-1} \)
Therefore, the rate of reaction is 2. \(14 × 10^{-4} Ms^{-1}\).