The formation of nitrogen dioxide from nitrogen dioxide and oxygen gas was studi

Question

The formation of nitrogen dioxide from nitrogen dioxide and oxygen gas was studied. The balanced equation for the reaction is: O_2(g) + 2 NO(g) rightarrow 2 NO_2(g) The chemist measured the concentration of the three gases at various time intervals. The data is in the table below. Construct a well labelled graph (labelled axis with units, a title, etc.) to represent this data. Along the y-axis plot gas concentration and time on the x-axis. What is the average rate of consumption of nitrogen oxide and oxygen over the entire 71 minute interval? Determine the average rate for each. What is the average rate of formation of nitrogen dioxide over the entire 71 minute interval? What is the average rate of the consumption of NO and O_2 and the production of NO_2 each, a) over the first 10 minutes b) over the last 10 minutes? Find the initial rates of consumption of O_2 and NO and initial rate of formation of NO_2 as well as the rates at 4 minutes and 41 minutes into the experiment.

Explanation / Answer

Q1.

averate rate in the 71 minute interval:

Average Rate = (Cfinal - Cinitial) / (Tfinal - Tinitial)

Average Rate = (0.000116- 0.000514) / (71- 0) = -0.000005605 mol of NO / L

for O2:

Average Rate = (Cfinal - Cinitial) / (Tfinal - Tinitial)

Average Rate = (0.000144- 0.000343) / (71- 0) = --0.00000280 mol of O2 / L

Q2.

formation of NO2:

Average Rate = (Cfinal - Cinitial) / (Tfinal - Tinitial)

Average Rate = (0.000399- 0) / (71- 0) =0.00000561971 mol of NO2 / L

Q3.

AVG. rate for:

t 10 min

for NO:

Average Rate = (Cfinal - Cinitial) / (Tfinal - Tinitial)

Average Rate = (0.000311- 0.000343) / (10- 0) = -0.0000032 mol of NO / L

for O2:

Average Rate = (Cfinal - Cinitial) / (Tfinal - Tinitial)

Average Rate = (0.000242- 0.000343) / (10- 0) = -0.0000101 mol of O2 / L

for NO2.

Average Rate = (Cfinal - Cinitial) / (Tfinal - Tinitial)

Average Rate = (0.000204- 0) / (10- 0) = 0.0000204mol of NO2 / L

for final 10 min:

for NO:

Average Rate = (Cfinal - Cinitial) / (Tfinal - Tinitial)

Average Rate = (0.000116- 0.000127) / (71- 51) = --5.5*10^-7 mol of NO / L

for O2:

Average Rate = (Cfinal - Cinitial) / (Tfinal - Tinitial)

Average Rate = (0.000144-0.000150) / (71- 51) = -3*10^-7 mol of O2 / L

for NO2.

Average Rate = (Cfinal - Cinitial) / (Tfinal - Tinitial)

Average Rate = (0.000399 - 0.000387) / (71-61) = 0.0000012 mol of NO2 / L

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