At 500 K in the presence of a copper surface, ethanol decomposes according to the following equation.
C2H5OH(g) CH3CHO(g) + H2(g)
The pressure of C2H5OH was measured as a function of time and the following data were obtained.
Time (s) PC2H5OH (torr)
0 250.
100. 237
200. 224
300. 211
400. 198
500. 185
Using the above data, deduce the rate law
a) Rate = k[C2H5OH]2
b) Rate = k[C2H5OH]
c) Rate = k[C2H5OH] / [H2]
d) Rate = k[C2H5OH] / [CH3CHO]
e) Rate = k
To deduce the rate law, we need to analyze the relationship between the rate of the reaction and the concentration of the reactant, which in this case is ethanol (C2H5OH).
First, let's calculate the change in concentration of ethanol over a given time interval. We'll use the data provided:
Δt (s) = 100
Δ[P(C2H5OH)] = 237 - 250 = -13 torr
Using the ideal gas law, we can convert the change in pressure to a change in moles:
Δn(C2H5OH) = Δ[P(C2H5OH)] * V / RT
Assuming the volume (V), gas constant (R), and temperature (T) are constant, we can consider them to be effectively canceled out, so:
Δn(C2H5OH) ~ Δ[P(C2H5OH)]
Now, let's compare the change in concentration of ethanol (C2H5OH) with the change in time (Δt).
Δt (s) = 100
Δ[C2H5OH] = (-13 torr) / Δt
Δ[C2H5OH] = -0.13 M/s
Now, let's consider the differences in concentration over two different time intervals and compare them:
Interval 1 (Δt = 100 s):
Δ[C2H5OH] = -0.13 M/s
Interval 2 (Δt = 100 s):
Δ[C2H5OH] = (-13 -(-13)) / 100 = 0 M/s
Since the concentration of ethanol does not change in Interval 2, we can conclude that the rate of the reaction is dependent on the square of the concentration of ethanol (C2H5OH), thus yielding the rate law:
Rate = k[C2H5OH]^2
Therefore, the correct answer is a) Rate = k[C2H5OH]^2.