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If the reaction is second order in a and first order in b, what is the rate when the initial concentration of [a] = 4.11 mol/l and that of [b] = 3.71 mol/l?

Sagot :

Answer:

New concentration of A =3 [A] Put the values we get new rate of reaction. − dtd[R]=k[3A] 2[B] Hence, the rate of reaction will increase by 9 times. (iii) Given that. New concentrations of A=2[A] New concentration of B=2[B] Plug the values we get.

Explanation:

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