##### 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

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