Question

The following data were collected for the decomposition of acetaldehyde, CH₃CHO (used in the manufacture of a variety of chemicals including perfumes, dyes, and plastics), into methane and carbon monoxide. The data was collected at a temperature at 535 °C.

CH₃CHO → CH₄ + CO

[CH3CHO] (mol L-1)	Time (s)
0.200	0
0.153	0.20 x 102
0.124	0.40 x 102
0.104	0.60 x 102
0.090	0.80 x 102
0.079	1.00 x 102
0.070	1.20 x 102
0.063	1.40 x 102
0.058	1.60 x 102
0.053	1.80 x 102
0.049	2.00 x 102

Make a graph of concentration versus time: this is the first step of half-life analysis.

QUESTIONS:

1. Use the graph to determine how long it takes for the CH₃CHO concentration to decrease from 0.200 mol L^-1 to 0.100 mol L^-1. Use two significant figures for the numerical value of time.
2. Use the graph to determine how long it takes for the CH₃CHO concentration to decrease from 0.100 mol L^-1 to 0.050 mol L^-1.

3. What is the order of the reaction?

Answer 1

The best way to identify fast and easy the rate of reaction AND the rate constant is via Graphical Method.

First, as the name implies, we need to graph all types of order (most common)

Zero = C vs t;

For zero order, there is no dependency of concentrations:

dC/dt = k*C^0

dC/dt = k

When developed:

C = C0 - kt

if x axis is "time" then the slope is "k", and y-intercept is initial concentration C0. y-axis if C (concentration)

First = ln(C) vs. t

For first order

dC/dt = k*C^1

dC/dt = k*C

When developed:

dC/C = k*dt

ln(C) = ln(C0) - kt

if x axis is "time" then the slope is "-k", and y-intercept is initial concentration C0. y-axis if ln(C) (natural logarithm of concentration)

Second = 1/C vs. t

For Second order

dC/dt = k*C^2

When developed:

dC/C^2 = k*dt

1/C= 1/C0 + kt

if x axis is "time" then the slope is "k", and y-intercept is initial concentration C0. y-axis if 1/(C (inverse of concentration)

Know, graph all data in the 3 graphical methods

Best fit is 2nd order

Q1

from 0.2 to 0.1 -->

k = slope = 0.0771

1/C = 1/C0 + k*t

1/(0.1) = 1/(0.2) +0.0771*t

t = (1/(0.1) - 1/(0.2))/0.0771

t = 64.85 s

Q2

from 0.1 to 0.05

1/(0.05) = 1/(0.10) +0.0771*t

t = (1/(0.05) - 1/(0.10) )/0.0771

t = 129.70 s

Q3

order:

129.70 /64.85 = 2

ratio is 2x, therefore, once again, it must be 2nd order