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For a reaction A + B ⟶ P, the rate is given by
Rate = k [A] [B]^{2}
How is the rate of reaction affected if the concentration of B is doubled?
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Solution 1
For a reaction, A + B⟶ P
Rate_{1} = k[A][B]^{2}
If the concentration of B is doubled
Rate_{2} = k[A][2B]^{2}
Rate_{1} = k[A][B]^{2}
Rate_{2} = k[A][2B]^{2}
Rate_{1} = B^{2}
Rate_{2} 4B^{2}
Rate_{2} = 4 Rate_{1}
The rate of reaction will be four times the initial rate.
Solution 2
A + B → P
Rate = k[A] [B]^{2}
Since the given reaction has order two with respect to reactant B, thus, if the concentration of B is doubled in the given reaction, then the rate of reaction will become four times.
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