Question

Use the table below to solve for the rate constant for the reaction A + 2B + 2C - products. Trial [A], M 0.420 1 [B],M 0.150 0.150 [C],M Initial rate, M/s 0.330 4.00 0.330 4.00 2 0.210 3 0.420 0.225 0.330 6.00 4 0.420 0.150 0.660 8.00 Type in a numerical answer only with the correct sig figs. Do NOT type in units.

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question: in....typed format...

$\Rightarrow$Answer:

Rate constant of the reaction: k = 192.4 M ^-1 .s ^-1( ...i.e. answer )

$\Rightarrow$Explanation:

Following is the complete Explanation, for the above Answer...

• Given:

We have been given the following information, in table format...

Trials	[A], M(mol/L)	[B],M (mol/L)	[C], M (mol/L)	initial rate, M/s (mol/L.s)
1	0.420	0.150	0.330	4.00
2	0.210	0.150	0.330	4.00
3	0.420	0.225	0.330	6.00
4	0.420	0.150	0.660	8.00

$\Rightarrow$balanced rxn:  A + 2B + 2C  $\rightarrow$ products   

• Step - 1:

​​​​​​​Let, the reaction rate law: of the given reaction, be the following:

$\Rightarrow$rate = k x [A] ^xx [B]^yx [C]^z   -------------------------------------------------Eq. ( 1 )

$\Rightarrow$Where:

1. k =  rate constant of the reaction.
2. [A] = Molar concentration of the reactant: 'A'. [=] mol/L
3. [B] = Molar concentration of the reactant 'B'. [=] mol/L
4. [C] = Molar conc. of the reactant 'C'. [=] mol/L
5. x = Order of the reaction, with respect to the reacatnt 'A'.
6. y = Order of the reaction, withb respect to the reactant 'B' .
7. z = Order of the reaction, with respect to the reactant 'C' .

• ​​​​​​​Step - 2:

​​​​​​​If we divide the rxn rate equations of: trial 1, by the trial 2, in Eq. ( 1 )  we will get the following:

$\Rightarrow$ initial rate1 / initial rate2 = k/ k  x ([0.420]/[0.210])^x x [0.150/0.150]^y x [0.330/0.330] ^z

$\Rightarrow$ [4.0 M/s / 4.0 M/s ] = 1 x ( 2.0 )^x x 1.0 x 1.0  

$\Rightarrow$ (2.0 ) ^x = 1.0

$\Rightarrow$  x x ln(2) = ln(1.0 )  

$\Rightarrow$x = ln(1.0) / ln(2.0) = 0.0 ( i.e. zero )

$\Rightarrow$Order of the reaction, w.r.t. reactant 'A' = zero (0.0 )

• Step - 3:

​​​​​​​​​​​​​​If we divide the rxn rate equations of: trial 3, by the trial 1, in Eq. ( 1 )  we will get the following:

$\Rightarrow$ initial rate3 / initial rate1 = k/ k  x ([0.420]/[0.420])^x x [0.225/0.150]^y x [0.330/0.330] ^z

$\Rightarrow$ [6.0 M/s / 4.0 M/s ] = 1 x ( 1.0 )^x x ( 1.5 ) ^y x ( 1.0 )^z

$\Rightarrow$ (1.5 )^y = 1.5  

$\Rightarrow$  y x ln(1.5) = ln(1.5)

$\Rightarrow$ y = ln(1.5) / ln(1.5) = 1.0  

$\Rightarrow$ Order of the reaction, w.r.t. reactant 'B' is one ( i.e.  1.0 ) ...

• Step - 4:

​​​​​​​​​​​​​​​​​​​​​If we divide the rxn rate equations of: trial 4, by the trial 1, in Eq. ( 1 )  we will get the following:

$\Rightarrow$ initial rate4 / initial rate1 = k/ k  x ([0.420]/[0.420])^x x [0.150/0.150]^y x [0.660/0.330] ^z

$\Rightarrow$ [8.0 M/s / 4.0 M/s ] = 1 x ( 1.0 )^x x ( 1.0 ) ^y x ( 2.0 )^z

$\Rightarrow$ (2.0 )^z = 2.0  

$\Rightarrow$  z x ln(2.0) = ln(2.0)

$\Rightarrow$ z = ln(2.0) / ln(2.0) = 1.0

$\Rightarrow$ Order of the reaction, w.r.t. reactant 'C' is one ( i.e.  1.0 ) ...

• Step - 5:

​​​​​​​Now let's plug-in values in Eq. ( 1 ), for the given: trial 1:

$\Rightarrow$rate = k x [A] ^xx [B]^yx [C]^z   -------------------------------------------------Eq. ( 1 )

$\Rightarrow$4.0 M/s =   k  x [0.420 M ] ⁰ x [0.150 M ] ^1.0 x [ 0.330 M ] ^1.0

$\Rightarrow$k = 192.4 M ^-1 .s ^-1 ( i.e. rate constant of the reaction )  

And from the unit, i.e. M^-1 .s^-1 of the rate constant, we can surely say that the given reaction is of the second order.

-------------------------------------------------------------------------------------------------------------

​​​​​​​