1) M M [ PCl 5 ] = M [ PCl 3 ] = M [ Cl 2 ] = M 2) The equilibrium constant, K c
ID: 1043277 • Letter: 1
Question
1)
M
M
[PCl5] = M [PCl3] = M [Cl2] =M
2) The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K.
H2(g) + I2(g) 2 HI (g)
Calculate the equilibrium concentrations of reactants and product when 0.212moles of H2and 0.212 moles of I2are introduced into a 1.00 L vessel at 698 K.
M
3)The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K.
PCl3(g) + Cl2(g) PCl5(g)
Calculate the equilibrium concentrations of reactant and products when 0.300moles of PCl3 and 0.300 moles of Cl2 are introduced into a 1.00 L vessel at 500 K.
Explanation / Answer
1)
PCl5(g) ----------------- PCl3(g) + CL2(g) Kc= 1.20x10^-2
0.358 0 0
-x +x +x
0.358-x +x +x
Kc= [PCL3][Cl2]/[PCl5]
1.20x10^-2 = x*x/(0.358-x)
for solving the equation
x=0.0598
at equilibrium
number of moles of PCl5 = 0.358-0.0598 =0.2982 moles
number of moles of PCl3 = 0.0598 moles
number of moles of Cl2= 0.0598 moles
volume = 1.00L
[PCl5] = 0.2982/1.00= 0.2982M
[PCl3]= 0.0598/1.0= 0.0598M
[Cl2]= 0.0598/1.0= 0.0598M
2)
H2(g) + I2(g) ------------------ 2 HI(g) Kc = 55.6
0.212 0.212 0
-x -x +2x
0.212-x 0.212-x +2x
Kc= [HI]^2/[H2][I2]
55.6 = (2x)^2/(0.212-x)(0.212-x)
55.6 = [2x/(0.212-x)]^2
for solving the equation
x= 0.167
number of moles of H2= 0.212-0.167=0.045 moles
number o fmoles of I2 = 0.212-0.167=0.045 moles
number o fmole sof HI= 2x= 2x0.167 =0.334 moles
[H2] = 0.045/1.0= 0.045M
[I2]= 0.045M
[HI] = 0.334/1.0= 0.334M
3)
PCl3(g) + Cl2(g) ---------------- PCl5(g)
0.300 0.3000 0
-x -x +x
0.300-x 0.300-x +x
Kc= [PCl5]/[PCl3][Cl2]
83.3= x*x/(0.300-x)
for solving the equation
x=0.299
[PCl3] 0.3-0.299=0.001M
[Cl2]== 0.3-0.299=0.001M
[PCl5] = 0.299M
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