#### Answer

$N_2$ and $H_2$ will decrease their concentrations, and $NH_3$ will increase its concentration.

#### Work Step by Step

1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ NH_3 ]^{ 2 }}{[ N_2 ][ H_2 ]^{ 3 }}$$
2. At equilibrium, these are the concentrations of each compound:
$[ N_2 ] = 0.60 \space M - x$
$[ H_2 ] = 0.76 \space M - 3x$
$[ NH_3 ] = 0.48 \space M+ 2x$
$$1.2 = \frac{(0.48 + 2x)^2}{(0.60 - x)(0.76-3x)^3}$$
Real solutions: x = 0.0144 and x = 1.027
1.027 cannot be the correct value, because if so: $[N_2] = 0.60 -1.027 =$ Negative number.
So: x = 0.0144
$[ N_2 ] = 0.60 \space M - 0.0144 = 0.59 M$
$[ H_2 ] = 0.76 \space M - 3(0.0144) = 0.72 M$
$[ NH_3 ] = 0.48 \space M+ 2(0.0144) = 0.051 M$