## Chemistry: Molecular Approach (4th Edition)

$$K_c = 27.2$$
1. Determine each molarity: $CO$ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 28.01 g/mol - Calculate the amount of moles: $$26.9 \space g \times \frac{1 \space mol}{ 28.01 \space g} = 0.960 \space mol$$ - Calculate the molarity: $$\frac{ 0.960 \space mol}{ 5.19 \space L} = 0.185 \space M$$ $H_2$ : ( 1.008 $\times$ 2 )= 2.016 g/mol - Calculate the amount of moles: $$2.34 \space g \times \frac{1 \space mol}{ 2.016 \space g} = 1.16 \space mol$$ - Calculate the molarity: $$\frac{ 1.16 \space mol}{ 5.19 \space L} = 0.224 \space M$$ $CH_3OH$ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 32.04 g/mol - Calculate the amount of moles: $$8.65 \space g \times \frac{1 \space mol}{ 32.04 \space g} = 0.270 \space mol$$ - Calculate the molarity: $$\frac{ 0.270 \space mol}{ 5.19 \space L} = 0.0520 \space M$$ 2. At equilibrium, these are the concentrations of each compound: $[ CO ] = 0.185 \space M - x$ $[ H_2 ] = 0.224 \space M - 2x$ $[ CH_3OH ] = 0 \space M + x$ 3. Using the concentration of $CH_3OH$ at equilibrium, find x: $0 + x = 0.0520$ $x = 0.0520$ $[ CO ] = 0.185 \space M - 0.0520 =0.133$ $[ H_2 ] = 0.224 \space M - 2*( 0.0520 ) = 0.120$ $[ CH_3OH ] = 0 \space M + 0.0520 =0.0520$ - The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ CH_3OH ]}{[ CO ][ H_2 ] ^{ 2 }}$$ 4. Substitute the values and calculate the constant value: $$K_C = \frac{( 0.0520 )}{( 0.133 )( 0.120 )^{ 2 }} = 27.2$$