Chp4_NO2N2O4 - The algebra of this was covered in a...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Practice Example 4.8 in more detail We examined in class the equilibrium for the reaction: 4 2 2 2 O N NO colorless brown reddish - . This reaction is particularly convenient for study in general chemistry as: (1) NO 2 has a reddish-brown color whereas N 2 O 4 is colorless; and (2) the equilibrium constants are neither so big nor so large near room temperature as to make the reaction insensitive to moderate temperature changes. In Practice Example 4.3, one obtained kJ H rxn 02 . 58 - = and one can determine that K 298 = 8.8 (from G f 0 data). Note the reaction is exothermic (and note below what happens to K as temperature increases for such an exothermic process). In this exercise, we seek to calculate the equilibrium constant at two other temperatures, 273K and 373K (i.e. 0ºC and 100ºC). What we are really exploring is the temperature dependence of the equilibrium constant
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . The algebra of this was covered in a previous file. The two key working equations are: R S RT H K ∆ + ∆-= ln and -∆ = 1 2 2 1 1 1 ln T T R H K K or -∆ =-1 2 2 1 1 1 ln ln T T R H K K The number crunching is not particularly taxing. One should discover (confirm this for practice) that: K 273 = 75.1 and K 298 = 8.8 and K 373 = 0.079 Note that as the temperature increases, K decreases and the equilibrium shifts left (favors ‘the stuff of the left’) favoring the NO 2 . Since the NO 2 is colored, an increase in temperature should deepen the color of the mixture of gases; conversely, a decrease in temperature should tend towards less color....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online