Question 5.39

[305693278]

In an experiment, 0.020 mol NO2 was introduced into a flask of volume 1.00 L and the reaction 2NO2(g) --- N2O4(g)was allowed to come to equilibrium at 298 K. (a) Using information in Table 5G.2, calculate the equilibrium concentrations of the two gases. (b) The volume of the flask is reduced to half its original value. Calculate the new equilibrium concentrations of the gases.

I was wondering why the new concentration of 2NO2 would be 0.04 M?

[605775499]

Consider PV=nRT
P=nRT/V
Pressure and volume are inversely proportional
Halving volume = doubling pressure

Consider P=n/V for a given temperature
(where n/V = concentration)
Doubling pressure = doubling concentration

Halving volume = doubling concentration. Original concentration = 0.02. New concentration 0.04.
o7

[Cara W]

For this problem, the molarities are given as #moles/ 1 liter. So, when the volume (1 liter) is halved, the molarity needs to be adjusted by dividing #moles/ 0.5 liters to get the new concentrations (which should be double the original values).

[Sydney Nguyen 1K]

First, the volume is halved, so now the volume of the flask is 0.50 L. The original mol of NO2 was 0.020 mol. To calculate the M for the new concentration, you divide mol/L. 0.020 mol/0.50 L is 0.04 M.