In torr pressure of C3H5OH? Rate equations?

Question

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
Since the pressure of a gas is directly proportional to the concentration of gas, we can express the rate law for a gaseous reaction in terms of partial pressures. Using the above data, deduce the rate law, the integrated rate law, and the value of the rate constant, all in terms of pressure units in atm and time in seconds. Predict the pressure of C2H5OH after 900. s from the start of the reaction. [Hint: To determine the order of the reaction with respect to C2H5OH, compare how the pressure of C2H5OH decreases with each time listing.]

Fern · Accepted Answer

A zero-order reaction has a constant rate that is independent of the reactant's concentrations. The rate law is as follows:    rate=k

237-224 torr/100 sec - 0.13 torr/sec

224-211 torr/100 sec = 0.13  torr/sec

Rate = k(C2H5OH)^0 = Rate = k

Integrated rate law =  [A]t = -kt +  [A]o

250 torr x  1 atm/760 torr = 0.329 atm

237 torr x 1 atm / 760 torr = 0.312 atm

0.312  atm = -k(100 sec)  +  0.329 atm

k = 0.329 – 0.312/ 100 sec =  1.70 x 10^-4 atm/second

After 900 sec:

[A]t = -1.70 x 10^-4 atm/second x 900   +  0.329 = -0.153 + 0.329 = 0.176 atm

0.176 atm x 760 torr/atm = 133 torr

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In torr pressure of C3H5OH? Rate equations?

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