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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