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Murljashka [212]
2 years ago
13

Compute Reaction Rates for All Seven Trials

Chemistry
2 answers:
jeka942 years ago
7 0

Answer : The reaction rate at 3^oC,24^oC,40^oC\text{ and }65^oC are 36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec

Solution : Given,

Mass of tablet = 1000 mg

Volume of water = 0.200 L

The given formula will be,

\text{Reaction rate}=\frac{\text{Mass of tablet}/\text{Volume of water}}{\text{Reaction time}}

Now we have to calculate the reaction rate at different temperatures and reaction time.

\text{Reaction rate at }3^oC=\frac{1000mg/0.200L}{138.5sec}=\frac{5000mg/L}{138.5sec}=36mg/L/sec

\text{Reaction rate at }24^oC=\frac{1000mg/0.200L}{34.2sec}=\frac{5000mg/L}{34.2sec}=142mg/L/sec

\text{Reaction rate at }40^oC=\frac{1000mg/0.200L}{26.3sec}=\frac{5000mg/L}{26.3sec}=190mg/L/sec

\text{Reaction rate at }65^oC=\frac{1000mg/0.200L}{14.2sec}=\frac{5000mg/L}{14.2sec}=352mg/L/sec

Therefore, the reaction rate at 3^oC,24^oC,40^oC\text{ and }65^oC are 36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec

Mila [183]2 years ago
5 0

Answer:

36, 146, 190, 352

Explanation:

That is the correct answer on edg

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Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit
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The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

brainly.com/question/9875355

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brainly.com/question/8918040

the ph of a solution

brainly.com/question/9560687

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