Question

Pls help me i really need help

Answer 1

13. t = 1183 min and 0.50 M.

14.
`t_(1/2)=2.67x10^(-8)hr`

Step-by-step explanation:

Hello!

13. In this case, according to the units, we infer this is a second-order reaction which has the following integrated rate law:

`(1)/([A]) =(1)/([A]_0) +kt`

Which can be solved for the time as shown below:

`t=( (1)/([A])-(1)/([A]_0))/(k)`

Thus, we plug in the given concentrations and rate constant to obtain:

`t=( (1)/(0.250M)-(1)/(0.850M))/(0.002387M^(-1)min^(-1))\\\\t= 1183min`

For the second part, we proceed by using the same rate constant and the new initial concentration as follows:

`(1)/([A]) =(1)/([A]_0) +kt\\\\(1)/([A]) =(1)/(0.750M) +0.680M^(-1)min^(-1)*0.996min\\\\(1)/([A]) =1.99,M`

`[A]=0.50M`

14. In this case, according to the units of the rate constant, we infer this is a zeroth-order reaction, therefore we compute the half-life has shown below:

`t_(1/2)=([A]_0)/(2k)`

Thus, we plug in to obtain:

`t_(1/2)=(2.696x10^(-6)M)/(2*50.5M*hr^(-1))`

`t_(1/2)=2.67x10^(-8)hr`

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