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

Answer:

a)

$K_2=1.3x10^{2}atm^{-1/2}$

b)

$PCl_5\rightarrow PCl_3+Cl_2$

$[PCl_5]=0.0375M$

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:

$K=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

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:

$[PCl_5]=\frac{[PCl_3][Cl_2]}{K}$

$[PCl_5]=\frac{\frac{0.20mol}{4L} *\frac{0.12mol}{4L} }{0.04}$

$[PCl_5]=0.0375M$

Best regards!