Based on your knowledge of chemical equations, what do you know for sure

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin

erastovalidia [21]

Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

8 0

3 years ago

PLEAEEEE HELPPP

alexgriva [62]

Speed is a scalar quantity( quantity having only magnitude). So it doesn't specifies direction. Hence option C.) 5Km/hr north is incorrect.

8 0

3 years ago

In the late 1400s,ties among people of the same formed the basics of most aspects of life in rural west Africa

lara31 [8.8K]

Can u write it more specifically? so that I can answer

6 0

3 years ago

Calculate the molarity 20ml of 3.5 m kci in to a final volume of 100 ml

satela [25.4K]

Answer:

Molarity = 0.7 M

Explanation:

Given data:

Volume of KCl = 20 mL ( 0.02 L)

Molarity = 3.5 M

Final volume = 100 mL (0.1 L)

Molarity in 100 mL = ?

Solution:

Molarity = number of moles of solute / volume in litter.

First of all we will determine the number of moles of KCl available.

Number of moles = molarity × volume in litter

Number of moles = 3.5 M × 0.02 L

Number of moles = 0.07 mol

Molarity in 100 mL.

Molarity = number of moles / volume in litter

Molarity = 0.07 mol /0.1 L

Molarity = 0.7 M

3 0

3 years ago

Water vapor changing to liquid water

kiruha [24]

Explanation:

When water vapour changes to liquid water then this process is known as condensation.

For example, when lid is placed in a hot water filled pan then after sometime vapours appear on the surface of lid. When temperature of water decreases then water vapours convert into liquid form.

Thus, we can conclude that in condensation water vapor changes to liquid water.

4 0

3 years ago

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