Experimental rate law, PLEASE HELP?
The experimental rate law for the reaction 3A + B -> 3C is
rate = k[A]^2[B].
What will happen to the reaction rate if the concentration of A is doubled?
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The experimental rate law for the reaction 3A + B -> 3C is
rate = k[A]^2[B].
What will happen to the reaction rate if the concentration of A is doubled?
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rate = k[A]^2[B]
If concentration of A is doubled then the reaction rate = k[2A]^2[B]
rate = 4k[A]^2[B]
Thus, the reaction rate will increase 4 times greater than the initial rate
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