A bar house with volume of 500 m^3, has 50 smokers in it. Each begins to smoke 2

Question

A bar house with volume of 500 m^3, has 50 smokers in it. Each begins to smoke 2 cigarettes per hour at time t=0 hr. An individual cigarette emits about 1.4 mg of formaldehyde (HCHO).Formaldehyde converts to carbon dioxide with a reaction rate coefficient k=0.4 /hr. No fresh enters the bar and no air exits the bar. Assume oxygen content is always sufficient (for human breath and cigarette combustion). The bar is a complete mixing environment with 1 atm of pressure at 25degrees C. If the threshold for people to feel eye irritation is 0.05 mg/m^3 of HCHO, how long will it take for people to feel that?

Explanation / Answer

*** Part 1 *** this is the "production" of CH2O part... CH2O is added at a constant rate of.. 50 smokers x (2 cigs/hr / smoker) x (1.4mg CH2O / cig) x (1g / 1000mg) x (1mol / 30.0g) = 4.67x10^-3 mol/hr and that is added to 500 m³ x (1000L / 1m³) = 5x10^5 L so.. the [CH2O] is ADDED at the rate (4.67x10-3 mol/hr) / 5x10^5 L) = 9.33x10^-5 M/hr *** part 2 *** rate [CH2O] is removed... you have this reaction.. (I think I'm reading that correctly) 1 CH2O + 1 O2 ---> 1 CO2 + 1 H2O assuming [O2] >>> [CH2O], the reaction is approximately zero order in O and 1st order in [CH2O] (I can tell that by the units.. zero order k has units "M / time", 1st order has units "1 / time", 2nd order has units "1 / (M x time)"... get it?).. so this is 1st order in [CH2O] so you have some rate equation for the COMBUSTION of formaldehyde like.. -d[CH2O] / dt = k x [CH2O]¹ or simply.. rate consumed = (0.4/hr) x [CH2O] *** part 3 *** the overall accumulation of CH2O.. which we can easily write as d[CH2O] / dt *** part 4 *** we need to write a "balance" on CH2O and here it is.. rate CH2O is ADDED - rate CH2O is consumed = rate CH2O is accumulated.. so that.. 9.33x10^-5 M/hr - (0.4/hr) x [CH2O] = d[CH2O] / dt *** 5 *** and finally, you need to solve that differential equation using these two data points.. (t, [CH2O]) = (0, 0) to.. (t, [CH2O]) = (t, 1.67x10^-9M) where the 2nd data point is from (0.05mg CH2O / m³) x (1g / 1000mg) x (1 mol / 30.0g) x (1m³ / 1000L) = 1.67x10^-9 M ie.. we start with [CH2O] = 0 at time = 0 and increase it to 1.67x10^-9M at time = t and solve for "t" ********** how are you with differential equations? can you finish? hint.. . d[CH2O] / dt = 9.33x10^-5 M/hr - (0.4/hr) x [CH2O] is of the form dy/dx = k - k'y or.. y' = k - k'y... where k is a constant.

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