Answer
$$K_p= 1.77$$
Work Step by Step
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 $K_p$
$$K_p = K_c(RT)^{2 - 3} = (2.46 \times 10^2)(0.08314 \times 1670)^{-1} = 1.77$$
