Question

For the following reaction, K_eq is 0.00183 at 390. K:

PCl₅(g)  PCl₃(g) + Cl₂(g)

If 3.72 g of PCl₅ is placed in a 3.68 L bulb at 390. K, what is the equilibrium pressure of Cl₂? 1 atm = 1.013 bar

bar

The number of significant digits is set to 3; the tolerance is +/-4%

Answer 1

$\bf \text{The no. moles of } PCl_5=\frac{3.72\;g}{208.24\;g/mol.}=0.018 \;mol$

$\bf \text{ Volume of container }=3.68 \;L$

PC15(8) =PC13(g) + Cl2(g) 0.018 0 0 с -X E 0.018 - x х х х х

Total no. of moles at equilibrium 0.018 + x

X The equilibrium pressure of chlorine Pcl, x (1 atm 0.018 + x

$\bf K_{eq}=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

$\bf 0.00183=\frac{(x/3.68L)(x/3.68L)}{(0.018-x)/3.68\;L}$

$\bf 0.00183=\frac{x^2}{(0.018-x)\times3.68}$

$\bf x^2=0.00183\times(0.018-x)\times3.68$

$\bf x^2=0.0067344\times(0.018-x)$

$\bf x^2+0.0067344x+0.000121=0$

$\bf x=0.00814$

Total no. of moles at equilibrium 0.018 + x

$\bf \text{ Total no. of moles at equilibrium }=0.018+0.00814=0.02614$

X The equilibrium pressure of chlorine Pcl, x (1 atm 0.018 + x

$\bf \text{ The equilibrium pressure of chlorine }P_{Cl_2}=\frac{0.00814}{0.02614}\times(1\;atm)$

$\large \bf \text{ The equilibrium pressure of chlorine }P_{Cl_2}=0.312\;atm$