Answer
$$K_p = 0.0249 = 2.49 \times 10^{-2}$$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ SCl_2 ]& [ C_2H_4 ]& [ S(CH_2CH_2Cl)_2 ]\\
Initial& 0.675 & 0.973 & 0 \\
Change& -x& -2 x& +x\\
Equilibrium& 0.675 -x& 0.973 -2 x& 0 +x\\
\end{vmatrix}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ SCl_2 ] = 0.675 \space M - x$
$ [ C_2H_4 ] = 0.973 \space M - 2x$
$ [ S(CH_2CH_2Cl)_2 ] = x$
3. Using the concentration of $ S(CH_2CH_2Cl)_2 $ at equilibrium, find x:
$x = 0.350 $
$ [ SCl_2 ] = 0.675 \space M - 0.350 =0.325 $
$ [ C_2H_4 ] = 0.973 \space M - 2*( 0.350 ) = 0.273 $
$ [ S(CH_2CH_2Cl)_2 ] =0.350 $
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ S(CH_2CH_2Cl)_2 ]}{[ SCl_2 ][ C_2H_4 ] ^{ 2 }}$$
4. Substitute the values and calculate the constant value:
$$K_C = \frac{( 0.350 )}{( 0.325 )( 0.273 )^{ 2 }} = 14.4$$
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1. Calculate $\Delta n$ (n is the amount of moles of gases):
$$\Delta n = n_{products} - n_{reactants} = 1 - 3 = -2 $$
2. Convert the temperature to Kelvin:
$$T/K = 20.0 + 273.15 = 293.15$$
3 . Calculate Kp:
$$K_p = K_c(RT)^{\Delta n} = ( 14.4 )(0.0821 \times 293.15 )^{ -2 }$$
$$K_p = 0.0249 $$
