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

$$K_p= 1.77$$
1. Determine the concentrations in mol/L: $H_2$ : ( 1.008 $\times$ 2 )= 2.016 g/mol - Calculate the amount of moles: $$1.00 \space g \times \frac{1 \space mol}{ 2.016 \space g} = 0.496 \space mol$$ - Calculate the molarity: $$\frac{ 0.496 \space mol}{ 0.500 \space L} = 0.992 \space M$$ $H_2S$ : ( 1.008 $\times$ 2 )+ ( 32.07 $\times$ 1 )= 34.09 g/mol - Calculate the amount of moles: $$1.06 \space g \times \frac{1 \space mol}{ 34.09 \space g} = 0.0311 \space mol$$ - Calculate the molarity: $$\frac{ 0.0311 \space mol}{ 0.500 \space L} = 0.0622 \space M$$ - For $S_2$ at equilibrium: $$\frac{ 8.00 \times 10^{-6} \space mol}{ 0.500 \space L} = 1.60 \times 10^{-5} \space M$$ 2. At equilibrium, these are the concentrations of each compound: $[ H_2 ] = 0.992 \space M - 2x$ $[ S_2 ] = 0 \space M - x$ $[ H_2S ] = 0.0622 \space M + 2x$ 3. Using the concentration of $S_2$ at equilibrium, find x: $0 - x = 1.60 \times 10^{-5}$ $x = -1.60 \times 10^{-5}$ $[ H_2 ] = 0.992 \space M - 2*( -1.60 \times 10^{-5} ) = 0.992$ $[ S_2 ] = 0 \space M - (-1.60 \times 10^{-5}) =1.60 \times 10^{-5}$ $[ H_2S ] = 0.0622 \space M + 2*( -1.60 \times 10^{-5} )= 0.0622$ 4. Write the expression and calculate $K_c$ $$K_c = \frac{[H_2S]^2}{[H_2]^2[S_2]} = \frac{(0.0622)^2}{(0.992)^2(1.60 \times 10^{-5})} =2.46 \times 10^2$$ 5. Calculate $K_p$ $$K_p = K_c(RT)^{2 - 3} = (2.46 \times 10^2)(0.08314 \times 1670)^{-1} = 1.77$$