Please show work and explain so I can understand exactly what needs to be done.
ID: 907387 • Letter: P
Question
Please show work and explain so I can understand exactly what needs to be done. Thank you for your help!
1) If a 1.00 L vessel contains 0.010 atm SO3, 0.010 atm O2, and 0.020 atm SO2, is the system at equilibrium with respect to the reaction below? If not, what will happen in order to establish equilibrium?
2 SO2 + O2 2 SO3 Kp = 0.14
2) Calculate the equilibrium concentrations of reactants and products of a 1.00 L vessel is charged with 0.100 mol each of H2 and I2.
H2 + I2 2. HI Kc = 54.4
3) Calculate the equilibrium concentrations of reactants and products in a 1.00 L vessel that was charged with 0.200 mol HI.
H2 + I2 2. HI Kc = 54.4
4) Calculate the equilibrium concentrations of reactants and products in a 1.00 L vessel that was charged with 0.010 mol H2, 0.020 mol I2, and 0.010 mol HI.
H2 + I2 2. HI Kc = 54.4
Explanation / Answer
1) 2 SO2 + O2 2 SO3 Kp = 0.14
Qp = Pso3^2/Pso2^2*PO2
= 0.01^2/(0.02^2*0.01) = 25
As Qp > Kp .the reaction shifts towards left side. to reach equilibrium.
2) H2 + I2 2 HI Kc = 54.4
initial 0.1 0.1 0
at equili 0.1-x 0.1-x 2x
Kc = [HI]^2/[H2][I2]
54.4 = 2X/((0.1-X)(0.1-X))
X = 0.055
concentration of [HI] = 2*0.055 = 0.11 M
[H2] = 0.1-0.055 = 0.045 M
[I2] = 0.1-0.055 = 0.045 M
3) H2 + I2 2 HI Kc = 54.4
initial 0.2 0.2 0
at equili 0.2-x 0.2-x 2x
Kc = [HI]^2/[H2][I2]
54.4 = 2X/((0.2-X)(0.2-X))
X = 0.13
concentration of [HI] = 2*0.13 = 0.26 M
[H2] = 0.2-0.13 = 0.07 M
[I2] = 0.2-0.13 = 0.07 M
4)
H2 + I2 2 HI Kc = 54.4
initial 0.01 0.02 0.01
at equili 0.01-x 0.02-x 0.01+2x
Kc = [HI]^2/[H2][I2]
54.4 = (0.01+2X )/((0.01-X)(0.02-X))
X = 0.000243
concentration of [HI] = 2*0.000243+0.01 = 0.010486 M
[H2] = 0.01-0.000243 = 0.009757 M
[I2] = 0.02-0.000243 = 0.019757 M
1) 2 SO2 + O2 2 SO3 Kp = 0.14
Qp = Pso3^2/Pso2^2*PO2
= 0.01^2/(0.02^2*0.01) = 25
As Qp > Kp .the reaction shifts towards left side. to reach equilibrium.
2) H2 + I2 2 HI Kc = 54.4
initial 0.1 0.1 0
at equili 0.1-x 0.1-x 2x
Kc = [HI]^2/[H2][I2]
54.4 = 2X/((0.1-X)(0.1-X))
X = 0.055
concentration of [HI] = 2*0.055 = 0.11 M
[H2] = 0.1-0.055 = 0.045 M
[I2] = 0.1-0.055 = 0.045 M
3) H2 + I2 2 HI Kc = 54.4
initial 0.2 0.2 0
at equili 0.2-x 0.2-x 2x
Kc = [HI]^2/[H2][I2]
54.4 = 2X/((0.2-X)(0.2-X))
X = 0.13
concentration of [HI] = 2*0.13 = 0.26 M
[H2] = 0.2-0.13 = 0.07 M
[I2] = 0.2-0.13 = 0.07 M
4)
H2 + I2 2 HI Kc = 54.4
initial 0.01 0.02 0.01
at equili 0.01-x 0.02-x 0.01+2x
Kc = [HI]^2/[H2][I2]
54.4 = (0.01+2X )/((0.01-X)(0.02-X))
X = 0.000243
concentration of [HI] = 2*0.000243+0.01 = 0.010486 M
[H2] = 0.01-0.000243 = 0.009757 M
[I2] = 0.02-0.000243 = 0.019757 M