Question

In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewhere you need to be in 3.5 hours and will not be able to monitor the reaction properly. What concentration of BC2 should you originally use in order to have 0.42 M remaining in 3.5 hours

Answer 1

Complete Question

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Answer:

The concentration of
`BC_2` that should used originally is
`C_Z_o = 0.4492M`

Step-by-step explanation:

From the question we are told that

The necessary elementary step is

`2BC_2 ----->4C + B_2`

The time taken for sixth of 0.5 M of reactant to react
`t = 9 hr`

The time available is
`t_a = 3.5 hr`

The desired concentration to remain
`C = 0.42M`

Let Z be the reactant , Y be the first product and X the second product

Generally the elementary rate law is mathematically as

`-r_Z = kC_Z^2 = - (d C_Z)/(dt)`

Where k is the rate constant ,
`C_Z` is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

For second order reaction

`(1)/(C_Z) - (1)/(C_Z_o) = kt`

Where
`C_Z_o` is the initial concentration of Z which a value of
`C_Z_o = 0.5M`

From the question we are told that it take 9 hours for the concentration of the reactant to become

`C_Z = C_Z_o - (1)/(6) C_Z_o`

`C_Z = 0.5 - (0.5)/(6)`

`= 0.4167 M`

So

`(1)/(0.4167) - (1)/(0.50) = 9 k`

`0.400 = 9 k`

=>
`k = 0.044\ L/ mol \cdot hr^(-1)`

For
`C_Z = 0.42M`

`(1)/(0.42) - (1)/(C_Z_o) = 3.5 * 0.044`

`2.38 - 0.154 = (1)/(C_Z_o)`

`2.226 = (1)/(C_Z_o)`

`C_Z_o = (1)/(2.226)`

`C_Z_o = 0.4492M`