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.
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
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: https://brainly.com/question/13314785
