For the reaction: 2A + B → A2B the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.
Answers
Answer:
Initial rate of reaction = 8 × 10^-9 mol/Ls
Rate of reaction when [A] is reduced to 0.06 mol/L = 3.89 × 10^-9 mol/Ls
Explanation:
The given reaction is
2A + B → A₂B
Given, initially
[A] = 0.1 mole/L
[B] = 0.2 mole/L
According to the question
The rate of reaction is given by
rate = k[A][B]², where k = 2.0 × 10^-6 L²/mol²s
Therefore,
Initial rate of reaction = 2.0 × 10^-6 × 0.1 × 0.2² = 8 × 10^-9 mol/Ls
When concentration of A is reduced to 0.06 mol/L, the concentration of A used in reaction = 0.1 - 0.06 = 0.04
Since two moles of A reacts with 1 mole of B
Therefore 0.04 moles of A will react with 0.02 moles of B
Therefore the concentration of B remaining will be 0.2 - 0.02 = 0.18
Now,
[A] = 0.06 mol/L
[B] = 0.18 mol/L
Therefore the rate of reaction = k[A][B]²
= 2.0 × 10^-6 × 0.06 × 0.18²
= 3.89 × 10^-9 mol/Ls