Question

The following results were obtained at 600 k for the decomposition of ethanol on an alumina (Al₂O₃) surface,

C₂H₅OH(g) → C₂H₄(g) + H₂O(g)

a. Predict P_total in torr at t=80s

b. What is the value of the rate constant, and what are its units?

c. what is the order of the reaction?

d. Calculate P_total at t = 300s.

Short answer

1. 370 torr
2. The units of k for a zero-order reaction are M/s, the units of k for a first-order reaction are 1/s, and the units of k for a second-order reaction are 1/(M·s).
3. Zero order
4. 500 torr

Step-by-step solution

Explanation for part (a)

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

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

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, thehalf-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 decompositionof 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.

a. we can use the equation and substitute the value of x as our time is 80s and calculate (P_total) as follows:

$$\begin{gathered} y=1.5x+250 \\ y=1.5\left(80\right)+250 \\ y=370torr \end{gathered}$$

Thus, the total pressure at 80s is 370 torr

Explanation for part (b)

b.The units of the rate constant, k, depend on the overall reaction order. The units of k for a zero-order reactionare M/s, the units of k for a first-order reaction are 1/s, and the units of k for a second-order reaction are 1/(M·s).

Explanation for part (c)

c. When we plot concentration vs. Time and get a straight line for any reaction, the reaction is zero order. Similarly in this case, the reaction iszero order.

Explanation for part (d)

d. we can use the equation and substitute the value of x as our time 300s and calculate total pressure as follows:

$$\begin{gathered} y=1.5x+250 \\ y=1.5\left(80\right)+250 \\ y=370torr \end{gathered}$$

However, this answer cannot be correct. Since we started out with a total pressure of 250 torr, the maximum pressure can reach only up to 500 torr once the reaction is complete. Thus, the total pressure at 300s is 500 torr