Many gaseous reactions occur in car engines and exhaust systems. One of these is

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

Solution:-

NO2(g) + CO(g) NO(g) + CO2(g)

Plots are as follows

NO2(g): 1) [NO2(g)] vs. time, =

2) ln [NO2(g)] vs. time, =

and 3 1/[NO2(g)] vs. time); Straight line means second order reaction

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.

This means reaction is second order with NO2 and zero order with CO

a) What are the reaction orders for each reactant?

Solution :- Order with NO2 = second order

And order with CO is zero order

b) What is the overall reaction order?

Solution :- Overall order of the reaction is the sum of the order of the individual reactants

So overall order = 2+0 = 2 that is second order

c) What is the rate law?

Solution :-

Rate law is as follows

Rate = K [NO2]^2

d) If you doubled the concentration of NO2(g) what would that do to the rate?

Solution :- Since the reaction is second order with the NO2 therefore if we double the concentration of the NO2 then rate of the reaction is quadruple means increase by factor 4

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)

Solution :-

Rate = K [NO2]^2

K = rate /[NO2]^2

K = 1.0*10^-5 mol L-1 s-1 / 0.10 mol L-1

K = 1.0*10^-4 M-1 s-1

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