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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 ^{o}C. Determine the rate law for this reaction, including the value for k.
NO(g) + H_{2}(g) ½N_{2}(g) + H_{2}O(g)
With initial P_{H2} = 400 mmHg |
With initial P_{NO} = 400 mmHg |
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Initial P_{NO}, mmHg |
Rate, mmHg/s |
Initial P_{H2}, 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}[H_{2}]^{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 H_{2} 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 H_{2}, use experiments 1 and 3 for the easiest numbers:
Within experimental error, n = 1, or first order in H_{2}.
The rate constant can now be found from the rate law: k = Rate/[NO]^{2}[H_{2}] for each experiment:
Initial P_{H2}, mmHg |
Initial P_{NO}, 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}