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For a reaction that follows the general rate law, rate = k[a][b]2, what will happen to the rate of reaction if the concentration of b is increased by a factor of 2.00? the rate will

User Khanzor
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1 Answer

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Answer:

If the concentration of b is increased by a factor of 2.00, the rate will be increased four times.

Step-by-step explanation:

This happens because the concentration of compound b is squared. This means that if the concentration doubles (2*b), by placing it in the equation, the coefficient will be squared, leaving 4*
b^(2).

In this way, the rate will increase 4 times concerning the original one.

rate1 = k[a][b]^2

rate2 = k[a][2*b]^2

rate2 = k[a]*4*[b]^2

rate2 = 4 * k[a][b]^2

rate2 = 4 * rate 1

Have a nice day!

User Neolaser
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