QuestionDetails: The following data was collected for the concentration ofNO 2 a

Question

QuestionDetails: The following data was collected for the concentration ofNO2 as a function of time: Time NO2(M) 0 1.00 10 0.951 20 0.904 30 0.860 0 1.00 10 0.951 20 0.904 30 0.860 40 0.818 50 0.778 60 0.740 70 0.704 40 0.818 50 0.778 60 0.740 70 0.704 80 0.670 90 0.637 100 0.606 80 0.670 90 0.637 100 0.606 1. What is the average rate of reaction between 10s and20s?
2. What is the average rate of the reaction between 50s and60s? 3. What is the rate of formation of O2 between 50sand 60s? Please any help would be appreciated and I promise to rate.Thanks 0 1.00 10 0.951 20 0.904 30 0.860

Explanation / Answer

'Average' rate is just the change in concentration over a certaintime interval. 1. Rate = [NO2] / time         = (0.904 - 0.951) M /(20 - 10) s         = -0.0047 M/s It's okay to have a negative rate - we are using change in reactantconcentration, so as time increases (as the reaction proceeds), theamount of reactant decreases, so the rate is negative. 2. Rate = (0.740 - 0.778)M / (60 - 50)s         = -0.0038 M/s 3. From the equation, we know that only half as much oxygen isproduced as there is nitrogen dioxide reacting. Therefore the rateof oxygen formation should be half the rate of NO2reaction. (i.e. it takes twice as long for a certain concentrationof oxygen to form, than it does for the same concentration ofNO2 to react). Rate of oxygen formation = -0.0038 / 2 = +0.0019 M/s (Don't forget to change the sign to +'ve, since oxygen is beingformed.)

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