What is the concentration of the base (NaOH) in this titration?

What is the rate law for the uncatalyzed reaction?

elena55 [62]

What is the rate law for the uncatalyzed reaction?

Metals often form several cations with different charges. Cerium, for example, forms and ions, and thallium forms and ions. Cerium and thallium ions react as follows:

2Ce^4+(aq) + Tl^+(aq) ---> 2Ce^3+(aq)+Tl^3+(aq)

This reaction is very slow and is thought to occur in a single elementary step. The reaction is catalyzed by the addition of Mn^2+ according to the following mechanism:

Ce^4+(aq)+Mn^2+(aq) ---> Ce^3+(aq)+Mn^3+(aq)

Ce^4+(aq)+Mn^3+(aq) ---> Ce^3+ (aq) + Mn^4+ (aq)

Mn^4+(aq) + Tl^+(aq) ---> Mn^2+(aq) + Tl^3+(aq)

If the uncatalyzed reaction occurs in a single elementary step, why is it a slow reaction?

1)The probability of an effective three-particle collision is low.

2) The transition state is low in energy.

3) The reaction requires the collision of three particles with the correct energy and orientation.

4) All reactions that occur in one step are slow.

The catalyzed reaction is first order in [Ce^4+] and first order in [Mn^2+]. Which of the steps in the catalyzed mechanism is rate determining?

Based on their available oxidation states, rank the following metals on their ability to catalyze this and other oxidation-reduction reactions.

Rank from best to worst catalyst.

Vanadium, titanium, manganese

4 0

3 years ago

Read 2 more answers

Refer to the experiment below:

saul85 [17]

Answer:

I think you are correct with your answer.

Explanation:

But I don't think that the explanation is correct because I think that it is the formation of both baking soda and water that makes it work...Vinegar doesn't make gas bubbles maybe because it hasn't reacted or maybe because it is not the right ingredient for the solution you are looking for. Thank you for the question.....Am not sure I really helped in your question but thank you..

8 0

3 years ago

Calculate the approximate enthalpy of the reaction in joules. Estimate that 1.0 mL of vinegar has the same thermal mass as 1.0 m

Nesterboy [21]

Explanation:

The given data is as follows.

Density of vinegar = 1.0 g/ml

Specific heat capacity = 4.25 $J/g ^{o}C$

$T_{1}$ = $17 ^{o}C$ , and $T_{2}$ = $14 ^{o}C$

Relation between enthalpy and specific heat is as follows.

$\Delta H = mC \Delta T$

Hence, putting the values into the above formula as follows.

$\Delta H = mC \Delta T$

= $25 \times 1.0 \times 4.25 J/g ^{o}C \times -3^{o}C$ (as density = $\frac{mass}{volume}$ )

= - 315 J

Thus, we can conclude that the enthalpy of reaction is -315 J.

As the value is negative so, it means that heat is releasing. Hence, the reaction is exothermic in nature.

5 0

3 years ago

What is the pH of a 0.025 M [OH] solution?

Debora [2.8K]

The answer is 12.4.I think its correct answer.

8 0

3 years ago

Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w

Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619= [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0

3 years ago