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? - Chemistry

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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

Rate1 = k[A][B]2

If the concentration of B is doubled

Rate2 = k[A][2B]2

Rate1 = k[A][B]2

Rate2 = k[A][2B]2

Rate1 = B2

Rate2 4B2

Rate2 = 4 Rate1

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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