Many gaseous reactions occur in car engines and exhaust systems. One of these is
ID: 879378 • Letter: M
Question
Many gaseous reactions occur in car engines and exhaust systems. One of these is
NO2(g) + CO(g) NO(g) + CO2(g)
An experiment is performed using a concentration of 0.10 mol L-1 for both NO2(g) and CO(g). The concentration of each reactant is monitored over time. Six separate plots, three for each reactant are generated (For NO2(g): 1) [NO2(g)] vs. time, 2) ln [NO2(g)] vs. time, and 3 1/[NO2(g)] vs. time); For CO(g): 4) [CO(g)] vs. time, 5) ln [CO(g)] vs. time, 6) and 1/[CO2(g)] vs. time). Plots 3 and 4 give straight lines, while all the other plots give curves.
a) What are the reaction orders for each reactant?
b) What is the overall reaction order?
c) What is the rate law?
d) If you doubled the concentration of NO2(g) what would that do to the rate?
e) If when using the concentrations above for each reactant (i.e. 0.10 mol L-1), you find the rate to be 1.0 x 10-5 mol L-1 s-1. Find k (don't forget to indicate the units for k).
Explanation / Answer
1) [NO2(g)] vs. time, : curve: May be seond order or first order
2) ln [NO2(g)] vs. time, curve: So Second order ( as for first order the graph will be straight line)
3 1/[NO2(g)] vs. time); straight line: second order
For CO(g):
4) [CO(g)] vs. time, : straight line : Zero order
5) ln [CO(g)] vs. time:curve
6) and 1/[CO2(g)] vs. time): curve
Plots 3 and 4 give straight lines, while all the other plots give curves.
a) What are the reaction orders for each reactant?
Answer: the order with respect to NO2 is second order and the order with respect to CO is zero order
b) What is the overall reaction order?
The overall reactio order will be two
c) What is the rate law?
Rate = k [NO2]2[CO]0
d) If you doubled the concentration of NO2(g) what would that do to the rate?
The rate will become four times if we increase the concentration of NO2 by two times.
e) Rate = rate constant X [NO2]2
1X 10^-5 = K X 0.1 X 0.1
K = 10^-7 mol-1 L second-1
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