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Question
For the reaction:
\[\ce{2A + B -> A2B}\]
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.
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Solution
Given: [A] = 0.1 mol L−1
[B] = 0.2 mol L−1
k = 2.0 × 10−6
The initial rate of the reaction is
Rate = k[A][B]2
= 2.0 × 10−6 × 0.1 × (0.2)2
= 8 × 10−9 mol L−1 s−1
When [A] reduces to 0.06 mol L−1, i.e. 0.04 mol L−1 of A has reacted, then the reactant B
= `1/2 xx 0.04`
= 0.02 mol L−1
Hence, the new [B] = 0.2 − 0.02 = 0.18 mol L−1
Thus, the new concentrations of A and B are
[A] = 0.06 mol L−1,
[B] = 0.18 mol L−1,
Now rate = 2.0 × 10−6 × (0.06) × (0.18)2
= 3.89 × 10−9 mol L−1 s−1
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