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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry, 4th edition.

Problem 13.48.

Listed below are initial rates, expressed in terms of the rate of decrease of partial pressure of a reactant for the following reaction at 826 oC. Determine the rate law for this reaction, including the value for k.

NO(g) + H2(g) ½N2(g) + H2O(g)

With initial

PH2 = 400 mmHg

With initial

PNO = 400 mmHg

Initial PNO,

mmHg

Rate,

mmHg/s

Initial PH2,

mmHg

Rate,

mmHg/s

359

0.750

289

0.800

300

0.515

205

0.550

152

0.125

147

0.395






Answer:

The unknown rate law is given by Rate = k[NO]m[H2]n. Using the Method of Initial rates will give the rate law and the value of the rate constant. Since the units cancel in the Method of Initial rates, we do not need to convert to molarity.

To find the order in NO, use the first set of data where the pressure of H2 is kept constant. Experiments 2 and 3 are about a factor of 2 difference in initial pressure of NO, so will give easier numbers to work with:

Within experimental error, m = 2, or second order in NO.

To find the order in H2, use experiments 1 and 3 for the easiest numbers:

Within experimental error, n = 1, or first order in H2.

The rate constant can now be found from the rate law: k = Rate/[NO]2[H2] for each experiment:

Initial PH2, mmHg

Initial PNO, mmHg

Initial rate, mmHg/s

k, mmHg–2 s–1

400

359

0.750

(0.750)/(400)(359)2

= 1.45×10–8

400

300

0.515

(0.515)/(400)(300)2

= 1.43×10–8

400

152

0.125

(0.125)/(400)(152)2

= 1.35×10–8

289

400

0.800

(0.800)/(289)(400)2

= 1.73×10–8

205

400

0.550

(0.550)/(205)(400)2

= 1.68×10–8

147

400

0.395

(0.395)/(147)(400)2

= 1.68×10–8

Averaging all of the k values gives k = 1.55×10–8 mmHg–2 s–1