Answer
The reverse reaction is favored, producing more solid $NH_4HS$.
Work Step by Step
1. Calculate the $Q_p$ for that reaction:
- The exponent of each concentration is equal to its balance coefficient.
$$Q_C = \frac{[Products]}{[Reactants]} = {[ NH_3 ][ H_2S ]}$$
2. Substitute the values and calculate $Q_c$:
$$Q_C = {( 0.166 )( 0.166 )} = 0.0276$$
$Q_c \gt K_c$, therefore, the reverse reaction is favored, producing more solid $NH_4HS$.
