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gizmo_the_mogwai [7]
2 years ago
6

The decomposition of ethanol (C2H5OH) on an alumina (Al2O3) surface C2H5OH1 g2 h C2H4 1 g2 1 H2O1 g2 was studied at 600 K. Conce

ntration versus time data were col- lected for this reaction, and a plot of [A] versus time resulted in a straight line with a slope of 24.00 3 1025 mol/L  s. a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. If the initial concentration of C2H5OH was 1.25 3 1022 M, calculate the half-life for this reaction. c. How much time is required for all the 1.25 3 1022 M C2H5OH to decompose?
Chemistry
1 answer:
Solnce55 [7]2 years ago
3 0

The rate constant for the decomposition of ethanol on an alumina surface is 24.00 × 10²⁵ M.

The rate law is: rate = 24.00 × 10²⁵ M/s

The integrated rate law is: [C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.

The time required for all the 1.25 × 10²² M C₂H₅OH to decompose is 5.21 × 10⁻⁵ s.

Let's consider the decomposition of ethanol on an alumina surface.

C₂H₅OH(g) ⇒ C₂H₄(g) + H₂O(g)

The plot of [A] vs time (t) resulted in a straight line, which indicates that the reaction follows zero-order kinetics.

The slope, 24.00 × 10²⁵ M/s, represents the rate constant, k.

<h3>What is zero-order kinetics?</h3>

It is a chemical reaction in which the rate of reaction is constant and independent of the concentration of the reacting substances

The rate law for zero-order kinetics is:

rate = 24.00 × 10²⁵ M/s

The integrated rate law for zero-order kinetics is:

[C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

<h3>What is the half-life?</h3>

Is the time for the amount of substance to decrease by half.

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, we can calculate the half-life [t(1/2)] using the following formula.

t(1/2) = [C₂H₅OH]₀ / 2 × k

t(1/2) = (1.25 × 10²² M) / 2 × (24.00 × 10²⁵ M/s) = 2.60 × 10⁻⁵ s

We can calculate the time required for all the 1.25 × 10²² M C₂H₅OH to decompose using the integrated rate law.

[C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

0 M = 1.25 × 10²² M - 24.00 × 10²⁵ M/s × t

t = 5.21 × 10⁻⁵ s

The rate constant for the decomposition of ethanol on an alumina surface is 24.00 × 10²⁵ M.

The rate law is: rate = 24.00 × 10²⁵ M/s

The integrated rate law is: [C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.

The time required for all the 1.25 × 10²² M C₂H₅OH to decompose is 5.21 × 10⁻⁵ s.

Learn more about zero-order kinetics here: brainly.com/question/13314785

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Calculate the standard entropy of vaporization of ethanol, C2H5OH, at 285.0 K, given that the molar heat capacity at constant pr
MArishka [77]

Answer: The standard entropy of vaporization of ethanol is 0.275 J/K

Explanation:

C_2H_5OH(l)\rightleftharpoons C_2H_5OH(g)

Using Gibbs Helmholtz equation:

\Delta G=\Delta H-T\Delta S

For a phase change, the reaction remains in equilibrium, thus \Delta G=0

\Delta H=T\Delta S

Given: Temperature = 285.0 K

\Delta H=78.3J/mol

Putting the values in the equation:

78.3J=285.0K\times \Delta S

\Delta S=0.275J/K

Thus  the standard entropy of vaporization of ethanol is 0.275 J/K

4 0
3 years ago
what is the concentration of hydroxide ions after 50.0 ml of 0.250 m naoh is added to 120 ml of 0.200 m na2so4? please show all
Galina-37 [17]

The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

What is meant by concentration?

Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.

Concentration of hydroxide ions can be calculated by,

M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.

where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.

Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

To learn more about concentration click on the given link brainly.com/question/17206790

#SPJ4

5 0
8 months ago
Please answer! I will mark you as brainlist! ​
blagie [28]

Answer:

a) no

b) no

c) yes

d) yes

Explanation:

7 0
2 years ago
If 5 grams of hydrogen reacted with 40 games of oxygen to form water, how much of water would be formed?
GREYUIT [131]
2H₂ + O₂ = 2H₂O

n(H₂)=m(H₂)/M(H₂)
n(H₂)=5g/2.0g/mol=2.5 mol

n(O₂)=m(O₂)/M(O₂)
n(O₂)=40g/32.0g/mol=1.25 mol

H₂ : O₂ = 2 : 1

2.5 : 1.25 = 2 : 1

n(H₂O)=n(H₂)=2n(O₂)=2.5 mol

m(H₂O)=n(H₂O)M(H₂O)

m(H₂O)=2.5mol*18.0g/mol=45.0 g
3 0
2 years ago
What is the mass number of an ion with 106 electrons, 157 neutrons, and a +1 charge?
il63 [147K]

Answer:

264 g/mol

Explanation:

#electrons equal #protons = 106

Plus 1 charge => m protons = 106 + 1 = 107

Mass number: 107 + 157 = 264 g/mol

5 0
3 years ago
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