Answer
The reaction proceeds to the left.
Work Step by Step
- $\Delta n = 2-2 = 0$
Therefore: $K_p = K_c$ and $Q_p = Q_c$
$K_c = 2.7$
Calculate $Q_c$:
1. The exponent of each concentration is equal to its balance coefficient.
$$Q_C = \frac{[Products]}{[Reactants]} = \frac{[ CO_2 ][ H_2 ]}{[ CO ][ H_2O ]} = \frac{\frac{n_{CO_2}}{V}\frac{n_{H_2}}{V}}{\frac{n_{CO}}{V}\frac{n_{H_2O}}{V}} = \frac{n_{CO_2}n_{H_2}}{n_{CO}n_{H_2O}} $$
2. Substitute the values and calculate the quotient value:
$$Q_C = \frac{( 0.62 )( 0.43 )}{( 0.13 )( 0.56 )} = 3.7$$
Since $Q_c \gt K_c$, the reaction will proceed to the left. (reactants side)
