Question

I have to take this quiz and I only have two questions left PLEASE HELP I DO NOT WANT TO FAIL!

1. For the following reaction, it is found that doubling the amount of A causes the reaction rate to double while doubling the amount of B causes the reaction rate to quadruple. What is the best rate law equation for this reaction?
A + B -> 2C
rate = k[A]2[B]
rate = k[A][B]
rate = k[A][B]2
rate = k[A]1/2[B]
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2. Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.

X + Y --> products

Trial [X] [Y] Rate
1 0.20 M 0.15 M 2.4 × 10-2 M/min
2 0.20 M 0.30 M 4.8 × 10-2 M/min
3 0.40 M 0.30 M 19.2 × 10-2 M/min

Answer 1

1. rate = k[A][B]² - is the best choice,
because when you double A. rate = k[2A][B]²=rate = k*2*[A][B]²

when you double B . rate = k[A][2B]²=rate = k[A][B]²*2²= rate = k[A][B]²*4
2.
1) rate1 =[0.20]^x*[0.15]^y=2.4*10⁻²
rate2=[0.20]^x*[0.30]^y=4.8*10⁻²
divide equation of rate2 by rate 1
rate2/rate1 [0.20]^x*[0.30]^y/([0.20]^x*[0.15]^y)=4.8*10⁻²/2.4*10⁻²

[0.30]^y/*[0.15]^y=4.8/2.4 , ( [0.30]/[0.15])^y=2, 2^y=1 y =1 , so exponent for [Y] will be 1

2)rate1= [0.20]^x*[0.30]^y = 4.8 × 10⁻²
rate2=[0.40]^x *[0.30 ]^y= 19.2 × 10⁻²
divide equation of rate2 by rate 1
rate2/rate1 [0.40]^x*[0.30]^y/([0.20]^x*[0.30]^y)=19.2*10⁻²/4.8*10⁻²
[0.40]^x/[0.20]^x=19.2/4.8
([0.40]/[0.20])^x= 4,
(2)^x=4, x=2, so so exponent for [X] will be 2

3) rate=k[X]²[Y]
4) to find k
take [X]=0.20 M, [Y]= 0.30 M rate=4.8 × 10⁻² M/min
rate=k[X]²[Y]
4.8 × 10⁻² M/min=k[0.20M]²[0.30M]
4.8 × 10⁻² M/min=k*(0.04*0.3)M³
k=(4.8 × 10⁻² M/min)/(0.012 M³)= 4 min/M²
5) final equation
rate=(4 min/M²)*k[X]²[Y]

Answer 2

Answer: 1.
`Rate=k[A]^1[B]^2`

2.
`Rate=k[X]^2[y]^1`

x= 2 , y= 1 Total order= 2+1= 3

rate constant (k)=
`4mol^(-2)L^(2)min^(-1)`

Explanation:-

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

`Rate=k[A]^x[B]^y`

k= rate constant

x = order with respect to A

y = order with respect to A

On doubling the concentration of A, rate doubles thus order w.r.t A is 1.

On doubling the concentration of B, rate quadruples thus order w.r.t B is 2.

Thus
`Rate=k[A]^1[B]^2`

(2)
`X+Y\rightarrow products`

`2.4* 10^(-2)=k[0.20]^x[0.15]^y` (1)

`4.8* 10^(-2)=k[0.20]^x[0.30]^y` (2)

Dividing 2 by 1

`(4.8* 10^(-2))/(2.4* 10^(-2))=(k[0.20]^x[0.30]^y)/(k[0.20]^x[0.15]^y)`

`2=2^y`

`2^1=2^y`

`y=1`

`4.8* 10^(-2)=k[0.20]^x[0.30]^y` (2)

`19.2* 10^(-2)=k[0.40]^x[0.30]^y` (3)

Dividing 3 by 2

`(19.2* 10^(-2))/(4.8* 10^(-2))=(k[0.40]^x[0.30]^y)/(k[0.20]^x[0.30]^y)`

`4=2^x`

`2^2=2^x`

`x=2`

rate law:
`Rate=k[X]^2[y]^1`

Total order = 2+1 = 3

using 1:
`2.4* 10^(-2)=k[0.20]^2[0.15]^1`

`k=4mol^(-2)L^(2)min^(-1)`