Question

At 25 deg. C, the second-order reaction NOCl(g)NO(g)+

1/2 CL2(g) is 50% complete after 5.82 hours when the
initialconcentration of NOCl is 4.46 mol/l. how long will it
takefor the reaction to be 75% complete?

Answer 1

The problem asks for the time when a second-order reaction of NOCl to NO and Cl2 will be 75% complete. Using the second-order integrated rate law, we calculate the time based on the initial concentration and the known rate constant.

Step-by-step explanation:

The question involves a second-order reaction involving NOCl turning into NO and Cl2. Given that the reaction is 50% complete after 5.82 hours with an initial concentration of 4.46 mol/L, we can use the integrated rate law for a second-order reaction to find how long it will take for the reaction to be 75% complete.

For a second-order reaction, the integrated rate law is given by:

1/[A] - 1/[A]0 = kt

where [A] is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is time. Since the reaction is 50% complete at t = 5.82 hrs, we can calculate the rate constant k using the initial concentration and the concentration at t = 5.82 hrs. Once we have the value of k, we can solve for t when the reaction is 75% complete.

If the reaction is 50% complete, the concentration of NOCl will be half of the initial concentration, which is 2.23 mol/L. We can then calculate k using the integrated rate law:

1/2.23 mol/L - 1/4.46 mol/L = k(5.82 hrs)

After finding k, we plug in the concentration that represents 75% completion, which is 1.115 mol/L (since at 75% completion, a quarter of the initial concentration 4.46 mol/L remains), and solve for t.

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Answer 2

It will take 17.46 hours for the reaction to be 75% complete.

Step-by-step explanation:

Half life for second order kinetics is given by:

`t_{(1)/(2)=(1)/(k* a_o)`

`t_{(1)/(2)` = half life = 5.82 hour

k = rate constant =?

`a_o` = initial concentration = 4.46 mol/L

`5.82 hour=(1)/(k* 4.46 mol/L)`

`k=(1)/(5.82 hour* 4.46 mol/L)=0.03852 L/mol hour`

Integrated rate law for second order kinetics is given by:

`(1)/([a])=kt+(1)/([a_o])`

[a] = concentration left after time t =
`100\%-75\%=25\% of [a_o]=0.25[a_o]`

`(1)/(0.25* 4.46 mol/L)=(1)/(0.03852 L/mol hour)* t+(1)/(4.46 mol/L)`

`t=17.46 hours`

It will take 17.46 hours for the reaction to be 75% complete.