0.1mol/1L = 0.1M
.........Si(s) + 2Cl2(g) ==> SiCl4(g)
initial..0......0.............0.1M
change....x......2x............-x
equil.....x......2x..........0.1-x
K = (SiCl4)/(Cl2)^2
Substitute from the ICE chart above and solve for x, then 2x for Cl2. Note that solid Si does not appear in the K expression.
At elevated temperatures, solid silicon reacts with chlorine gas to form gaseous SiCl4. At some temperature, the equilibrium constant for this reaction is 0.30. If the reaction is started with 0.10 mol of SiCl4 in a one-liter flask, how much Cl2 will be present when equilibrium is established?
3 answers
ICE techniques
Si(s)+ 2Cl2(g) >> SiCl4(g)
Keq= [SiCl4]/[Cl2]^2
Cl2 SiCl4
2x .1-x concentrations
.3=(.1-x)/(2x)^2
1.2 x^2+x-.1=0
x= (-1+- sqrt (1+.48))/2.4 check that quadratic equation.
x= (-1+sqrt(1.48))/2.4=.09
concentration of Cl2= 2x=.18moles/liter
moles in one liter flask= .18
check all this math.
Si(s)+ 2Cl2(g) >> SiCl4(g)
Keq= [SiCl4]/[Cl2]^2
Cl2 SiCl4
2x .1-x concentrations
.3=(.1-x)/(2x)^2
1.2 x^2+x-.1=0
x= (-1+- sqrt (1+.48))/2.4 check that quadratic equation.
x= (-1+sqrt(1.48))/2.4=.09
concentration of Cl2= 2x=.18moles/liter
moles in one liter flask= .18
check all this math.
thank you both