1. ## chemistry work

1. In a closed container, nitrogen and hydrogen react to produce ammonia. the equilibrium constant is 0.050. at a specific time in the process $\displaystyle [NH_2(g)]$ is $\displaystyle 2.0 * 10^{-4}$ mol/L, $\displaystyle [H_2(g)]$ is $\displaystyle 4.0 * 10^{-3}$ mol/L and $\displaystyle [NH_3(g)]$is $\displaystyle 2.2 * 10^{-4} mol/L$

$\displaystyle N_2(g)$ + $\displaystyle 3H_2(g)$ forward reverse $\displaystyle 2NH_3(g)$

In which direction must the reaction produce to establish equilibrium?

I thought;

$\displaystyle 0.050= \frac{[NH_3)]^2}{[N_2(g)][H_2(g)]^3}$

$\displaystyle 0.050$$\displaystyle =\frac{[2.2*10^{-4}]^2}{[2.0*10^-4][4.0*10^-3]^3}$

doesen't seem to be getting me anywhere tho

2. Consider the system

$\displaystyle CO_2(g)$ + $\displaystyle H_2(g)$ forward reverse $\displaystyle CO(g)$ + $\displaystyle H_2O(g)$

Initially, 0.25 mol of water vapor and 0.20 mol of carbon monoxide are placed in a 1.00 L reaction vessel. At equilibrium, spectroscopic evidence shows that 0.10 mol of carbon dioxide is present. calculate K for the reaction.

2. This equation will be true at equilibrium. To make this equation true, will you need to increase the numbers in the numerator or denominator? Which way must the reaction proceed to accomplish this?

2. Consider the system

+ forward reverse +

Initially, 0.25 mol of water vapor and 0.20 mol of carbon monoxide are placed in a 1.00 L reaction vessel. At equilibrium, spectroscopic evidence shows that 0.10 mol of carbon dioxide is present. calculate K for the reaction.
Since one $\displaystyle CO$ and one $\displaystyle H_2O$ form one $\displaystyle CO_2$ and one $\displaystyle H_2$, 0.10mol of $\displaystyle CO$ and 0.10mol of $\displaystyle H_2O$ form 0.10mol of $\displaystyle CO_2$ and 0.10mol of $\displaystyle H_2$
so
amount of $\displaystyle H_2$ = 0.10
amount of $\displaystyle CO$ = 0.20-0.10
$\displaystyle H_2O$ = 0.25-0.10

$\displaystyle K = \frac{[CO][H_2O]}{[CO_2][H_2]}$