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 $[NH_2(g)]$ is $2.0 * 10^{-4}$ mol/L, $[H_2(g)]$ is $4.0 * 10^{-3}$ mol/L and $[NH_3(g)]$is $2.2 * 10^{-4} mol/L$

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

In which direction must the reaction produce to establish equilibrium?

I thought;

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

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

doesen't seem to be getting me anywhere tho

2. Consider the system

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

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 $CO$ and one $H_2O$ form one $CO_2$ and one $H_2$, 0.10mol of $CO$ and 0.10mol of $H_2O$ form 0.10mol of $CO_2$ and 0.10mol of $H_2$
so
amount of $H_2$ = 0.10
amount of $
CO$
= 0.20-0.10
$H_2O$ = 0.25-0.10

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