Question

Consider the following reactions and their equilibrium constants at 303 K. PCl5(g) equilibrium reaction arrow 1/4 P4(g) + 5/2 Cl2(g); Kc = 1.18 ✕ 10−21 1/4 P4(g) + 3/2 Cl2(g) equilibrium reaction arrow PCl3(g); Kc = 2.28 ✕ 1026 Calculate Kc for PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) at the same temperature.

Answer 1

Answer: The value of
`K_c` for the net reaction is
`2.69* 10^(5)`

Step-by-step explanation:

The given chemical equations follows:

Equation 1:
`PCl_5(g)\xrightarrow[]{K_1} (1)/(4)P_4(g)+(5)/(2)Cl_2(g)`

Equation 2:
`(1)/(4)P_4(g)+(3)/(2)Cl_2(g)\xrightarrow[]{K_2} PCl_5(g)`

The net equation follows:

`PCl_5(g)\xrightarrow[]{K_c} PCl_3(g)+Cl_2(g)`

As, the net reaction is the result of the addition of two above equations. So, the equilibrium constant for the net reaction will be the multiplication of two above equations.

The value of equilibrium constant for net reaction is:

`K_c=K_1* K_2`

We are given:

`K_1=1.18* 10^(-21)`

`K_2=2.28* 10^(26)`

Putting values in above equation, we get:

`K_c=1.18* 10^(-21)* 2.28* 10^(26)=2.69* 10^(5)`

Hence, the value of
`K_c` for the net reaction is
`2.69* 10^(5)`