## Chemistry: Molecular Approach (4th Edition)

$$K_c = 764$$
1. Find each molarity. $H_2$ : ( 1.008 $\times$ 2 )= 2.016 g/mol - Calculate the amount of moles: $$0.763 \space g \times \frac{1 \space mol}{ 2.016 \space g} = 0.378 \space mol$$ - Calculate the molarity: $$\frac{ 0.378 \space mol}{ 3.67 \space L} = 0.103 \space M$$ $I_2$ : ( 126.9 $\times$ 2 )= 253.8 g/mol - Calculate the amount of moles: $$96.9 \space g \times \frac{1 \space mol}{ 253.8 \space g} = 0.382 \space mol$$ - Calculate the molarity: $$\frac{ 0.382 \space mol}{ 3.67 \space L} = 0.104 \space M$$ $HI$ : ( 1.008 $\times$ 1 )+ ( 126.9 $\times$ 1 )= 127.9 g/mol - Calculate the amount of moles: $$90.4 \space g \times \frac{1 \space mol}{ 127.9 \space g} = 0.707 \space mol$$ - Calculate the molarity: $$\frac{ 0.707 \space mol}{ 3.67 \space L} = 0.193 \space M$$ 2. At equilibrium, these are the concentrations of each compound: $[ H_2 ] = 0.103 \space M - x$ $[ I_2 ] = 0.104 \space M - x$ $[ HI ] = 0 \space M + 2x$ 3. Using the concentration of $HI$ at equilibrium, find x: $0 + 2 x = 0.193$ $2 x = 0.193$ $x = \frac{ 0.193 }{ 2 }$ $x = 0.0965$ $[ H_2 ] = 0.103 \space M - 0.0965 =6.50 \times 10^{-3}$ $[ I_2 ] = 0.104 \space M - 0.0965 =7.50 \times 10^{-3}$ $[ HI ] = 0 \space M + 2*( 0.0965 )=0.193$ - The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ HI ] ^{ 2 }}{[ H_2 ][ I_2 ]}$$ 4. Substitute the values and calculate the constant value: $$K_C = \frac{( 0.193 )^{ 2 }}{( 6.50 \times 10^{-3} )( 7.50 \times 10^{-3} )} = 764$$