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Chem 1310: Introduction to physical chemistry Part 2: exercises 13-16, p658 Decomposition of N 2 O 5
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2 N 2 O 5 4 NO 2 + O 2 Given data for concentrations vs time, calculate average rates. To avoid confusion, first convert to standard units (s instead of h for time). tt[N 2 O 5 ] (hr)(s)(mol/L) 0.0000.849 0.5018000.733 1.0036000.633 2.0072000.472 3.00108000.352 4.00144000.262 5.00180000.196
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Decomposition of N 2 O 5 Calculating average rates. tt[N 2 O 5 ] (hr)(s)(mol/L) 0.0000.849 0.5018000.733 1.0036000.633 2.0072000.472 3.00108000.352 4.00144000.262 5.00180000.196 av(0..1800) = (0.849-0.733)/(1800-0) = 6.44*10 -5 mol L -1 s -1 av(1800..3600) = (0.733-0.633)/(3600-1800) = 5.56*10 -5 mol L -1 s -1 av(7200..18000) = (0.633-0.196)/(18000-7200) = 2.56*10 -5 mol L -1 s -1
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Decomposition of N 2 O 5 Obtaining the rate law tt[N 2 O 5 ]av rate (hr)(s)(mol/L)(mol L -1 s -1 ) 0.0000.849 6.44E-05 0.5018000.733 5.56E-05 1.0036000.633 4.47E-05 2.0072000.472 3.33E-05 3.00108000.352 2.50E-05 4.00144000.262 1.83E-05 5.00180000.196 Looks linear, i.e. first-order in [N 2 O 5 ]: rate = k [N 2 O 5 ]
![Decomposition of N 2 O 5 Obtaining the rate law tt[N 2 O 5 ]av rate (hr)(s)(mol/L)(mol L -1 s -1 ) E E E E E E Looks linear, i.e.](http://images.slideplayer.com./16/4964217/slides/slide_4.jpg "first-order in [N 2 O 5 ]: rate = k [N 2 O 5 ].")
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Decomposition of N 2 O 5 Obtaining the rate law (2) tt[N 2 O 5 ]av rate (hr)(s)(mol/L)(mol L -1 s -1 ) 0.0000.849 6.44E-05 0.5018000.733 5.56E-05 1.0036000.633 4.47E-05 2.0072000.472 3.33E-05 3.00108000.352 2.50E-05 4.00144000.262 1.83E-05 5.00180000.196 Or test individual values: rate 0 /rate 7200 = 1.93 [N 2 O 5 ] 0 /[N 2 O 5 ] 7200 = 1.80 rate 3600 /rate 14400 = 2.44 [N 2 O 5 ] 3600 /[N 2 O 5 ] 14400 = 2.41 Not perfect but acceptable.
![Decomposition of N 2 O 5 Obtaining the rate law (2) tt[N 2 O 5 ]av rate (hr)(s)(mol/L)(mol L -1 s -1 ) E E E E E E Or test individual values: rate 0 /rate 7200 = 1.93 [N 2 O 5 ] 0 /[N 2 O 5 ] 7200 = 1.80 rate 3600 /rate = 2.44 [N 2 O 5 ] 3600 /[N 2 O 5 ] = 2.41 Not perfect but acceptable.](http://images.slideplayer.com./16/4964217/slides/slide_5.jpg "Decomposition of N 2 O 5 Obtaining the rate law (2) tt[N 2 O 5 ]av rate (hr)(s)(mol/L)(mol L -1 s -1 ) E E E E E E Or test individual values: rate 0 /rate 7200 = 1.93 [N 2 O 5 ] 0 /[N 2 O 5 ] 7200 = 1.80 rate 3600 /rate = 2.44 [N 2 O 5 ] 3600 /[N 2 O 5 ] = 2.41 Not perfect but acceptable.")
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Decomposition of N 2 O 5 Obtaining the rate constant tt[N 2 O 5 ]av rateest k (hr)(s)(mol/L)(mol L -1 s -1 )(s -1 ) 0.0000.849 6.44E-057.59E-05 0.5018000.733 5.56E-057.58E-05 1.0036000.633 4.47E-057.07E-05 2.0072000.472 3.33E-057.06E-05 3.00108000.352 2.50E-057.10E-05 4.00144000.262 1.83E-057.00E-05 5.00180000.196 k av = 7.23*10 -5 s -1 = 0.260 h -1
![Decomposition of N 2 O 5 Obtaining the rate constant tt[N 2 O 5 ]av rateest k (hr)(s)(mol/L)(mol L -1 s -1 )(s -1 ) E E E E E E E E E E E E k av = 7.23*10 -5 s -1 = h -1](http://images.slideplayer.com./16/4964217/slides/slide_6.jpg "Decomposition of N 2 O 5 Obtaining the rate constant tt[N 2 O 5 ]av rateest k (hr)(s)(mol/L)(mol L -1 s -1 )(s -1 ) E E E E E E E E E E E E k av = 7.23*10 -5 s -1 = h -1")
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Decomposition of N 2 O 5 Averaging over large periods, compare with "exact rate" tt[N 2 O 5 ]av ratefrom (hr)(s)(mol/L)(mol L -1 s -1 )rate law 0.0000.849 0.5018000.733 1.0036000.633 2.0072000.472 3.33E-053.44E-053.31E-052.98E-05 3.00108000.352 4.00144000.262 5.00180000.196 using [N 2 O 5 ] = (0.472+0.352)/2 = 0.412 mol/L But what about the concentrations we used when calculating the rate constant?
![Decomposition of N 2 O 5 Averaging over large periods, compare with exact rate tt[N 2 O 5 ]av ratefrom (hr)(s)(mol/L)(mol L -1 s -1 )rate law E E E E using [N 2 O 5 ] = ( )/2 = mol/L But what about the concentrations we used when calculating the rate constant](http://images.slideplayer.com./16/4964217/slides/slide_7.jpg "Decomposition of N 2 O 5 Averaging over large periods, compare with exact rate tt[N 2 O 5 ]av ratefrom (hr)(s)(mol/L)(mol L -1 s -1 )rate law E E E E using [N 2 O 5 ] = ( )/2 = mol/L But what about the concentrations we used when calculating the rate constant")
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Decomposition of N 2 O 5 Obtaining the rate constant with averaged concentrations - better precision tt[N 2 O 5 ]av rateest k (hr)(s)(mol/L)(mol L -1 s -1 )(s -1 ) 0.0000.849 0.7916.44E-058.15E-05 0.5018000.733 0.6835.56E-058.13E-05 1.0036000.633 0.5534.47E-058.09E-05 2.0072000.472 0.4123.33E-058.09E-05 3.00108000.352 0.3072.50E-058.14E-05 4.00144000.262 0.2291.83E-058.01E-05 5.00180000.196 k av = 8.10*10 -5 s -1 (c.f. 7.23*10 -5 s -1 found earlier)
![Decomposition of N 2 O 5 Obtaining the rate constant with averaged concentrations - better precision tt[N 2 O 5 ]av rateest k (hr)(s)(mol/L)(mol L -1 s -1 )(s -1 ) E E E E E E E E E E E E k av = 8.10*10 -5 s -1 (c.f.](http://images.slideplayer.com./16/4964217/slides/slide_8.jpg "7.23*10 -5 s -1 found earlier).")
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Decomposition of N 2 O 5 How do all concentrations vary with time?
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