## General Chemistry: Principles and Modern Applications (10th Edition)

(a) Since $Q_c \lt K_c$, the mixture is not at equilibrium; (b) And the net change will occur in the products direction.
1. Let's call the equal mass 'm', then the number of moles of each compound is: $$n_{compound}= \frac{m}{M(compound)}$$ 2. Calculate each molar mass: $CO$ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 28.01 g/mol $H_2O$ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol $CO_2$ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol $H_2$ : ( 1.008 $\times$ 2 )= 2.016 g/mol 3. Write the Q expression for this reaction, where the equilibrium does not depend on the volume: $$Q = \frac{n_{CO_2}n_{H_2}}{n_{CO}n_{H_2O}} = \frac{ \frac{m}{44.01} \frac{m}{2.016} }{ \frac{m}{28.01} \frac{m}{18.02}} = \frac{ \frac{1}{44.01} \frac{1}{2.016} }{ \frac{1}{28.01} \frac{1}{18.02}} = 5.69$$ Since $Q_c \lt K_c$, the mixture is not at equilibrium, and the net change will occur in the products direction, in order to increase $Q_c$.