The `K_(c)` for the equilirbium `2CO_(2(g)) Leftrightarrow 2CO_((g))+O_(2(g))" is "6.4 xx 10^(-7)` Predict whether reaction will take place to the left or to the right to reach equilibrium or remains or equilibrium
(a) `[CO_(2)]=5.3 xx 10^(-2), [CO]=3.6 xx 10^(-4), [O_(2)]=2.4 xx 10^(-3)`
(b) `[CO_(2)]=1.78 xx 10^(-1), [CO]=2.1 xx 10^(-2), [O_(2)]=5.7 xx 10^(-5)`
(c) `[CO_(2)]=1.03 xx 10^(-1), [CO]=2.4 xx 10^(-2), [O_(2)]=1.18 xx 10^(-5)`

Explanation

To solve the problem, we will calculate the reaction quotient \( Q_c \) for each case and compare it with the equilibrium constant \( K_c \) to determine the direction of the reaction.

### Given:
- The equilibrium reaction:
\[ 2 \text{CO}_2(g) \leftrightarrow 2 \text{CO}(g) + \text{O}_2(g) \]
- The equilibrium constant:
\[ K_c = 6.4 \times 10^{-7} \]

...