## Equilibrium Composition

Anna Nordstrom 1A
Posts: 15
Joined: Fri Sep 25, 2015 3:00 am

### Equilibrium Composition

On page 125 of our course reader the example problem, it says .150 mole of NH3 is formed but the concentration of N2 is decreased by half that amount. The constants are 1N2+ 3H2=2NH3 so I know the numbers had to be adjusted for the molar ratios but I am still a little unsure how.

Thanks!

Matthew_2C
Posts: 3
Joined: Fri Sep 25, 2015 3:00 am

### Re: Equilibrium Composition

N2 is decreased by half the amount that NH3 is increased because their molar ratio is 1 to 2. This means that for any amount of moles of N2 that is reacted with H2, twice that amount of moles of NH3 will be created.

Bryan Nguyen 1A
Posts: 20
Joined: Fri Sep 25, 2015 3:00 am

### Re: Equilibrium Composition

You can also see it as $\frac{0.150 mol NH_{3}}{L} * \frac{1 mol N_{2}}{2 mol NH_{3}}$ if you like to see how the concept works in an equation, and the same thing can be used for H2, where you would replace 1 mol N2 with 3 mol H2, so the concentration of H2 would be 1.5 times as much as the concentration of NH3.