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ArbitrLikvidat [17]
3 years ago
12

The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism

s for this reaction are: Mechanism A (1) H2(g)+I2(g)−→k12HI(g) (one-step reaction) Mechanism B (1) I2(g)⥫⥬=k−1k12I(g) (fast, equilibrium) (2) H2(g)+2I(g)−→k22HI(g) (slow) Mechanism C (1) I2(g)⥫⥬=k−1k12I(g) (fast, equilibrium) (2) I(g)+H2(g)−→k2HI(g)+H(g) (slow) (3) H(g)+I(g)−→k3HI(g) (fast) Which of these mechanisms are consistent with the observed rate law? mechanism C mechanism B mechanism A In 1967, J. H. Sullivan showed that this reaction was dramatically catalyzed by light when the energy of the light was sufficient to break the I−I bond in an I2 molecule. Which mechanism or mechanisms are consistent with both the rate law and this additional observation? mechanism A mechanism B mechanism C
Chemistry
1 answer:
tia_tia [17]3 years ago
3 0

Answer:

The only mechanism consistent with observed rate law is mechanism A.

Mechanism A<em> </em>is the only consistent with the observation of J. H. Sullivan

Explanation:

In a reaction, rate law is determined by slow step of the reaction.

In mechanism A, rate law is:

rate = k₁ [H₂][I₂]

In mechanism B, rate law is:

rate = k₂ [H₂][I]²

In mechanism C, rate law is:

rate = k₂ [H₂][I]

Thus, the only mechanism consistent with observed rate law is <em>mechanism A</em>

<em></em>

Now, catalyst works in the slow-step of reaction, is the reaction is catalyzed when The I-I bond breaks, the mechanism consistent with the observation is the mechanism where the slow-step of reaction involves this bond break, again, <em>mechanism A </em>is the only consistent with the observation of J. H. Sullivan.

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002 (part 1 of 2) 10.0 points
BARSIC [14]

Answer:

4) 1.5 mol

Explanation:

Well, the equation is already balanced and the mole to mole ratio of reactants and products are all 1. So if the limiting reactant is HCl and you have 1.5 mol, you do the mole to mole ratio with NaCl and since it is 1 to 1, there'd be 1.5 mol of NaCl.

6 0
2 years ago
A chemist has some 40% acid solution, some 60% acid solution, and a wholebunch of free time. How many liters of each should be u
TiliK225 [7]

Answer:

30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.

Explanation:

Let volume of the 40% acid solution be x.

Let volume of the 60% acid solution be y.

Volume of solution formed after mixing both solution = 40 L

x + y = 40 L..[1]

Volume of acid 40% solution = 40% of x= 0.4x

Volume of acid 60% solution = 60% of y= 0.6y

Volume of acid formed = 45% of 40 L = \frac{45}{100}\times 40=18L

0.4x+0.6y=18 L..[2]

Solving [1] and [2]

x = 30 L  ,   y = 10 L

30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.

8 0
3 years ago
(chem) which is more concentrated: 45.0 grams of HCOOH dissolved in 189 mL of water or 1.5 moles of CH↓2COOH dissolved in twice
Liula [17]

Answer:

CH3COOH would be more concentrated

Explanation:

The higher the concentration value, the more concentrated it is.

The relationship between concentration, moles and volume is given by the equation;

Concentration = No of moles / Volume

5.0 grams of HCOOH dissolved in 189 mL of water

Number of moles = Mass / Molar mass = 5 / 46.03 = 0.1086 mol

Concentration = 0.1086 / 0.189 = 0.5746 mol/L

1.5 moles of CH3COOH dissolved in twice as much water

Volume = 2 * 189 = 378 ml = 0.378 L

Concentration = 1.5 / 0.378 = 3.9683 mol/L

Comparing both concentration values;

CH3COOH would be more concentrated

6 0
2 years ago
A solution has a poh of 7. 1 at 10∘c. what is the ph of the solution given that kw=2. 93×10−15 at this temperature? remember to
Gnom [1K]

A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

pOH of solution = 7.1

Kw =2.93×10^(-15)

Firstly, we will calculate the value of pKw

The expression which we used to calculate the pKw is,

pKw=-log [Kw]

Now by putting the value of Kw in this expression,

pKw =−log{2.93×10^(-15)}

pKw =15log(2.93)

pKw=14.5

Now we have to calculate the pH of the solution.

As we know that,

pH+pOH=pKw

Now put all the given values in this formula,

pH+7.1=14.5

pH=7.4

Therefore, we find the value of pH of the solution is, 7.4.

learn more about pH value:

brainly.com/question/12942138

#SPJ4

7 0
1 year ago
A gas is located inside a closed container. if the gas is allowed to occupy more volume inside the container, what will happen t
nataly862011 [7]
Pressure in the container will decrease.
7 0
3 years ago
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