Chem 1310: Introduction to physical chemistry Part 2: exercises 13-16, p658 Decomposition of N 2 O 5.

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Chem 1310: Introduction to physical chemistry Part 2: exercises 13-16, p658 Decomposition of N 2 O 5

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)

Decomposition of N 2 O 5 Calculating average rates. tt[N 2 O 5 ] (hr)(s)(mol/L) av( ) = ( )/(1800-0) = 6.44*10 -5 mol L -1 s -1 av( ) = ( )/( ) = 5.56*10 -5 mol L -1 s -1 av( ) = ( )/( ) = 2.56*10 -5 mol L -1 s -1

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. first-order in [N 2 O 5 ]: rate = k [N 2 O 5 ]

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.

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

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?

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. 7.23*10 -5 s -1 found earlier)

Decomposition of N 2 O 5 How do all concentrations vary with time?