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

Question

7. Many gaseous reactions occur in car engines and exhaust systems. One of these is NO (g) +CO(g) NO(g) + CO2(g) An experiment is performed using a concentration of 0.10 mol L-1 for both NO(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) [NO (g)] vs. time, 2) In [NO (g)] vs. time, and 3 1/[NO(g)] CO(g): 4) [CO(g)] vs. time, 5) In [CO(g)] vs. time, 6) and 1/ICO2(g)] vs. time). Plots 3 and 4 give straight lines, while all the other plots give curves. vs. time; For 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), you find the rate to be 1.0 x 103 mol L's. Find k (don't forget to indicate the units for k).

Explanation / Answer

Sol. If the plot of 1/[Reactant] vs time gives a straight line , then the reaction is second order with respect to that reactant .

And If the plot of [Reactant] vs time gives a straight line , then the reaction is zero order with respect to that reactant .

Now , in the question ,

a) Plot no. 3 , that is , 1/[NO2(g)] vs time gives a straight line , so , reaction is second order with respect to NO2(g) .

Also , Plot no. 4 , that is , [CO(g)] vs time gives a straight line, so , reaction is zero order with respect to CO(g) .

b) Overall reaction order = 2+0 = 2 .

c) Rate law = k [NO2(g)]2 [CO(g)]0   = k [NO2(g)]2 where k = rate constant

d) if we double the conc. of NO2(g) , then , Rate law

= k ( 2 × [NO2(g)] )2   = 22 × k [NO2(g)]2 = 4 × k [NO2(g)]2 .So , rate law would become 4 times the original value .

e) As Rate = k [NO2(g)]2  so , K = Rate/[NO2(g)]2

k = 1.0 × 10-5   / (0.10)2   = 10-3   mol-1  L s-1   .

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