Question

1. The gases SO2, O2 and SO3 are allowed to reach equilibrium at a constant temperature. The equilibrium constant for the reaction

2SO2(g) + O2(g) -> 2S03(g)
is 1.6 x 104 atm-1

a) Calculate the value of Kp for the reaction
SO2(g) + ½O2(g) -> SO3(g)

(b) The equilibrium constant for the dissociation of Pcl5(g) to form PCl3(g) and Cl2(g) is
0.04 at 250°C. An equilibrium mixture contains 0.20 mol PC13 and 0.12 mol Cl, in a
4000 cm container.
i) Write the chemical equation.
ii) Calculate the concentration of PCIs in this container.
​

Answer 1

a)

`K_2=1.3x10^(2)atm^(-1/2)`

b)

`PCl_5\rightarrow PCl_3+Cl_2`

`[PCl_5]=0.0375M`

Step-by-step explanation:

Hello!

a) In this case, since we can see that the second reaction is equal to the half of the first reaction, we can relate the equilibrium constants as shown below:

`K_2=K_1^(1/2)`

Thus, by plugging in the the equilibrium constant of the first reaction we obtain:

`K_2=(1.6x10^4atm^(-1))^(1/2)\\\\K_2=1.3x10^(2)atm^(-1/2)`

b) In this case, for the described reaction we can write:

`PCl_5\rightarrow PCl_3+Cl_2`

Thus, the corresponding equilibrium expression is:

`K=([PCl_3][Cl_2])/([PCl_5])`

In such a way, since we know the equilibrium constant and the concentrations of PCl3 and Cl2 at equilibrium, we can compute the concentration of PCl5 at equilibrium as follows:

`[PCl_5]=([PCl_3][Cl_2])/(K)\\`

`[PCl_5]=((0.20mol)/(4L) *(0.12mol)/(4L) )/(0.04)`

`[PCl_5]=0.0375M`

Best regards!