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.