Chemistry, 22.06.2019 10:10 babyphoraaaaa

For the reaction, 4 a(g) + 3 b(g) => 2 c(g), the following data were obtained at constant temperature. experiment initial[a],mol/l initial [b],mol/l initial rate, m/min 1 0.200 0.150 5.00 2 0.400 0.150 10.0 3 0.200 0.300 10.0 4 0.400 0.300 20.0 which of the following is the correct rate law for the reaction? 1. rate = k[a]2[b]2 2. rate = k[a][b] 3. rate = k[a]2[b] 4. rate = k[a][b]2

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Answers

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 from: andrewschmitz132

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) $X+Y\rightarrow products$

$2.4\times 10^{-2}=k[0.20]^x[0.15]^y$ (1)

$4.8\times 10^{-2}=k[0.20]^x[0.30]^y$ (2)

Dividing 2 by 1

$\frac{4.8\times 10^{-2}}{2.4\times 10^{-2}}=\frac{k[0.20]^x[0.30]^y}{k[0.20]^x[0.15]^y}$

2=2^y

2^1=2^y

y=1

$4.8\times 10^{-2}=k[0.20]^x[0.30]^y$ (2)

$19.2\times 10^{-2}=k[0.40]^x[0.30]^y$ (3)

Dividing 3 by 2

$\frac{19.2\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.40]^x[0.30]^y}{k[0.20]^x[0.30]^y}$

4=2^x

2^2=2^x

x=2

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

Total order = 2+1 = 3

using 1: $2.4\times 10^{-2}=k[0.20]^2[0.15]^1$

$k=4mol^{-2}L^{2}min^{-1}$

Answer from: pronces

The reaction is first order overall.

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

For the given rate law:

Rate=k[A]^1[B]^1

x = 1= order with respect to A

y =1= order with respect to B

n =( x+y)= (1+1) = 2 = Total order

a) If [A] is doubled, the reaction rate will increase by a factor of 2: True

Rate'=k[2A]^1[B]^1

Rate'=k[2]^1[A]^1[B]^1

$Rate'=[2]^1\times Rate$

$Rate'=[2]\times Rate$

b) The reaction is first order overall: False

The overall order is 1+1= 2.

c) k is the reaction rate constant: True

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

k= rate constant

d) The reaction is first order in [B]: True

For the given rate law:

Rate=k[A]^1[B]^1

x = 1= order with respect to A

e) The reaction is first order in [A]: True

For the given rate law:

Rate=k[A]^1[B]^1

y =1= order with respect to B

Answer from: shyanne9364

C: AB is an intermediate; therefore the rate law is determined from the reactants in the previous step.

Explanation: : Rate law is usually determined from the slowest step of the reaction mechanism.

Thus as we can see that the reaction which is slow has AB as its intermediate, rate law can be determined from the same.

Moreover , we can see that the overall rate law is very different from the rat law determined from the slowest step.

Answer from: ejohnstonee111

Explanation:

To solve this question, we just need to put the new number into the equation. If [A] remain constant then that mean [A2]= [A1]. If B doubled, then that mean [B2]= 2[B2]. To find what factor does the rate of reaction increases, we need to divide the first reaction rate with the second. The calculation will be:

rate2/rate1= k[A2][B2]³ / k[A1][B1]³

rate2/rate1= [A1][2B1]³ / [A1][B1]³

rate2/rate1= A1*8B1³ / A1*B1³

rate2/rate1= 8/1= 8

The rate of reaction will be 8 times faster.

Answer from: needhelpwithHW10

Answer : The correct option is, (B) $M^{-1}s^{-1}$

Explanation:

Rate of reaction : It is defined as the rate of change in concentration of reactant or product with respect to time.

The given rate expression is,

Rate=k[A][B]

From this expression we conclude that the power of concentration of reactant A and B are 1, 1. That means it is a second order reaction.

The formula for determining the unit of 'k' is:

$\text{Unit}=\frac{(Concentration)^{1-n}}{Time}$

where, n = order of reaction

The unit of concentration is, M or mole/L

The unit of time is, second or 's'.

When n = 0

The unit of 'k' = $\frac{(M)^{1-0}}{s}=Ms^{-1}$

When n = 1

The unit of 'k' = $\frac{(M)^{1-1}}{s}=s^{-1}$

When n = 2

The unit of 'k' = $\frac{(M)^{1-2}}{s}=M^{-1}s^{-1}$

and so on.....

Therefore, the unit of 'k' of the following expression is, $M^{-1}s^{-1}$

Answer from: ash2905

Answer :

Step 1 : $A+B\rightleftharpoons AB(fast)\\Rate=K[A][B]$

Step 2 : $AB+B\rightarrow AB_2(slow)\\Rate=K[AB][B]$

Step 3 : $AB_2+B\rightleftharpoons AB_3(fast)\\Rate=K[AB_2][B]$

overall reaction rate : Rate=K[A][B]^2

As we know that the slow step is the rate determining step of the chemical reaction that determines the rate at which overall reaction proceeds. But in the slow step reaction, intermediate AB is involved. As, the intermediate will get completely consumed in the reaction. Hence, the rate law for the overall equation is not the same as the rate equation for the rate-determining step.

To determining the overall reaction rate following steps are involved :

From the step 2, the rate is,

Rate=K[AB][B] ...........(A)

Now we have to determine the equilibrium expression from the step 1 reaction.

$A+B\rightleftharpoons AB$

$K=\frac{[AB]}{[A][B]}$

[AB]=K[A][B]

Now put the value of [AB] in equation (A), we get

Rate=KK[A][B][B]

KK become new constant K'

Rate=K'[A][B]^2

Thus, this is the overall rate of reaction.

Answer from: ayoismeisalex

B. Rate = k[A]

order of reaction = 1

k = 0.213 s-1

Explanation:

Hello,

In this case, you should consider the experiments and how the concentration of both A and B affect the rate, thus, we can make some conclusions:

- Between the experiment 1 and 2, we can notice that modifying the concentration of B does not affect the rate as it remains in 3.20 x 10-1 M/s, for that reason the reaction is zeroth-order respect to B.

- Between the experiment 2 and 3, we can notice that doubling the concentration of A from 1.50 M to 3.00 M results in a doubling of the rate from 3.20 x 10-1 M/s to 6.40 x 10-1 M/s, for that reason, we can infer that the reaction is first-order respect to A.

In such a way, we can infer that the rate law is:

$Rate=k[A]^1[B]^0\\\\Rate= k[A]$

Thus, the order of reaction is first-order and the rate constant turns out:

$k=\frac{Rate}{[A]}=\frac{3.2x10^{-1}M/s}{1.50M}\\ \\k=0.213s^{-1}$

Therefore, answer is:

B. Rate = k[A]

order of reaction = 1

k = 0.213 s-1

Regards.

Answer from: LTLICKME7437

In this equation, [A] and [B] express the concentrations of A and B

Answer from: AlaskaAirlines

rate= k[A]²[B]²[C]

Explanation:

When concentration of A is increased two times ,keeping other's concentration constant , rate of reaction becomes 4 times .

So rate is proportional to [A]²

When concentration of B is increased two times , keeping other's concentration constant,rate of reaction becomes 4 times.

So rate is proportional to [B]²

When concentration of C is increased two times , keeping other's concentration constant, rate of reaction becomes 2 times.

So rate is proportional to [C]

So rate= k[A]²[B]²[C]

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