Question

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

Answer 1

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.

What is zero-order kinetics?

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

What is the half-life?

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