The following initial rate data are for the oxidation of nitrogen monoxide by ox
ID: 304616 • Letter: T
Question
The following initial rate data are for the oxidation of nitrogen monoxide by oxygen at 25 °C: 2 NO +02+2 NO,2 [Nolo, M 8.42x103 1.68x102 8.42x10-3 1.68x102 Experiment 02lo, M 6.01x10 01x10 1.20x103 1.20% 10-3 Initial Rate, Ms 3.58x104 1.42x10-3 7.15x104 2.84x10 .3 Complete the rate law for this reaction in the box below. Use the form kIAl B", where 'I' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n. Rate- From these data, the rate constant is ?",-1Explanation / Answer
rate = K [A]^m * [B]^n
rate = K [ NO] ^m * [O2]^n
for 1 and 2 exp the [O2] kept constant
rate1 = K [NO]1^m * [ O2]^n
rate 2 = K [NO]2^m * [O2]^n
rate2 / rate1 = ([NO]2/[NO]1)^m
1.42*10^-3 /3.58*10^-4= (1.68*10^-2 /8.42*10^-3)^m
4 = ( 2)^m
so here m = 2
similarly for 1 and 3 the [NO] is constant
7.15*10^-4 / 3.58*10^-4 = (1.2*10^-3 / 6.01/10^-4 )^n
2 =( 2)^1
so here n = 1
rate law is
rate = K [ NO]^2 * [ O2 ]
3.58*10^-4 = K * (8.42*10^-3)^2 * (6.01*10^-4)
K = 3.58*10^-4 / 0.426*10^-7
K = 8.404 *10^3 L2 mol-2s-1
similarly for the second problem
rate = K [NH4+]^m * [NO2-]^n
for 1 and 2 exp the [NH4+] is constant
6.94*10^-5 / 3.47*10^-5 = (0.310 / 0.155 )^n
2 = (2)^n
so here n = 1
similarly for 1 and 3 the [NO2-] is constant
6.92*10^-5 / 3.47*10^-5 = (1.32 / 0.662 )^m
2 = (2)^m
so here m = 1
rate law is
rate = K [ NH4+ ] *[ NO2-]
3.47*10^-5 = K (0.662)*(0.155)
K = 3.47*10^-5 / 0.103
K = 3.369*10-4 M-1 S-1
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