1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anika [276]
3 years ago
15

Consider the following mechanism: (1) ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast] (2) I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [s

low] (3) OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast] (a) What is the overall equation? Select the single best answer. ClO−(aq) + I−(aq) → IO−(aq) + H2O(l) + Cl−(aq) ClO−(aq) + I−(aq) ⇌ IO−(aq) + Cl−(aq) ClO−(aq) + I−(aq) ⇌ IO−(aq) + H2O(l) + Cl−(aq) ClO−(aq) + I−(aq) → IO−(aq) + Cl−(aq) (b) Identify the intermediate(s), if any. Select the single best answer. No intermediates Cl−, OH−, I−, ClO−, IO− HClO, OH−, HIO HClO, OH−, HIO, H2O (c) What are the molecularity and the rate law for each step? Select the single best answers. (1): bimolecular unimolecular termolecular rate = k1([HClO][OH−])/([HClO][OH−]) k1[HClO][OH−] k1[ClO−][H2O] (2): bimolecular unimolecular termolecular rate = k2[HIO][Cl−] k2([HIO][Cl−])/([I−][HClO]) k2[I−][HClO] (3): bimolecular unimolecular termolecular rate = k3[OH−][HIO] k3([H2O][IO−])/([OH−][HIO]) k3[H2O][IO−] (d) Is the mechanism consistent with the actual rate law: rate = k[ClO−][I−]? no yes
Chemistry
1 answer:
mash [69]3 years ago
8 0

Answer:

1. The overall equation is ClO-(aq)+I-(aq) → Cl-(aq)+IO-(aq)

2. The intermediates include: HClO(aq), OH-(aq) and HIO(aq)

3. The rates are k[ClO-][H2O], k[I-][HClO] and k[OH-][HIO]

4. No,

The rate depends on [OH-], so it's not consistent with the actual rate law

Explanation:

1

Given

(1) ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast]

(2) I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow]

(3) OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]

Add up the three equations

ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast]

I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow]

OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]

Remove all common terms {H2O(l) + I-(aq) +HClO(aq) +OH-(aq) +HIO(aq) => HClO(aq) + OH-(aq) + HIO(aq)

+H2O(l)}

We're left with

ClO-(aq) + I-(aq) => Cl-(aq) + IO-(aq)

2.

There are intermediates generated but they are not visible in the overall equation.

The intermediates include: HClO(aq), OH-(aq) and HIO(aq)

3.

The three steps are bimolecular.

The rates are k[ClO-][H2O], k[I-][HClO] and k[OH-][HIO]

4. Let K represents equilibrium constant

At step 1,

K1 = [HClO][OH-]/[ClO-]

Simplify;

K1 [ClO-]= [HClO][OH-]

K1[ClO-]/[OH-] = [HClO]

Determine the rate at step 2

= k2[I-][HClO]

= K1k2[I-][ClO-]/[OH-]

= k[ClO-][I-]/[OH-]

The answer is no

You might be interested in
In a neutralization reaction between sodium hydroxide,naoh and hydrochloric acid,hcl,the salt produced would be
natta225 [31]
NaOH + HCl ----> NaCl + H2O

3 0
3 years ago
Which answer choice describes the noble gases?​
Svetradugi [14.3K]

Answer:

18...

HE

NE

AR

KR

XE

RN

OG....................

3 0
3 years ago
 What is the subject chemistry actually all about? Is it hard, and should I take regular or honors?
horsena [70]
It depends if you are taking the class in high school or college. For college doing the experiments is fun but the calculations is not. Also if you are taking it, bring friends.
7 0
3 years ago
Read 2 more answers
What is the IUPAC name of KNO3<br>​
Tasya [4]
Answer: potassium nitrate


hope this helps
4 0
3 years ago
Read 2 more answers
A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
Colt1911 [192]

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

7 0
3 years ago
Other questions:
  • What is the molarity of a solution containing 23 g of NaCl in 500 mL of solution?
    14·1 answer
  • What is Pangaea?
    15·1 answer
  • Enter your answer in the provided box. phosphorus pentachloride is used in the industrial preparation of organic phosphorus comp
    9·1 answer
  • carbon has three main adjectives 12C 13c and 14c by comparing the atomic structure of the isotopes explain what is meant by the
    5·1 answer
  • 1) A certain liquid has a density of 0.80 g/ml. What is the mass of a 40 ml sample of this liquid? 1)
    11·1 answer
  • Sadiq repeated the experiment by adding sulphuric acid to magnesium carbonate. Write the word equation for this reaction below:
    12·1 answer
  • What is a chromatogram?<br>please help me out <br>Thank you<br>​
    7·2 answers
  • Transform boundaries occur where two plates – a slide past each other. b slide apart from each other. c slide towards each other
    10·1 answer
  • 2. If the half-life of Francium-233 is 22 minutes, how much time must elapse in order for 60 grams of a 64 gram
    11·1 answer
  • Na podstawie położenia w u.o. oraz liczb A i Z:
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!