Question

The equilibrium constant, K_c, for the following reaction is 41.1 at 289 K.

2CH₂Cl₂(g) --> CH₄(g) + CCl₄(g)  

When a sufficiently large sample of CH₂Cl₂(g) is introduced into an evacuated vessel at 289 K, the equilibrium concentration of CCl₄(g) is found to be 0.202 M.  

Calculate the concentration of CH₂Cl₂ in the equilibrium mixture. ? M

2)The equilibrium constant, K_c, for the following reaction is 1.80×10^-2 at 698 K.  
Calculate K_p for this reaction at this temperature.  

2HI(g) --> H₂(g) +  I₂(g)

3) The equilibrium constant, K_c, for the following reaction is 5.10×10^-6 at 548 K.  
Calculate K_p for this reaction at this temperature.  

NH₄Cl(s) --> NH₃(g) +  HCl(g)

Answer 1

Let us write ICE table for the equilibrium reaction

                                                       2CH₂Cl₂(g) --> CH₄(g)     +          CCl₄(g)

                     I                                     a moles                 0           0

                     C                                   -x moles               +x/2                   +x/2

                      E                                    a-x                       x/2                    x/2

Note: It is assumed that before equilibrium reached, x moles of CH2Cl2 dissociates.

Since, the Stocihiometric ratio between the reactant and the products is 2:1:1, x moles of the recctant will produce x/2 moles of each product.

Equilibrium constant for the reaction

Kc =        [CH4] x [CCl4]     ------------------------------------------- Eq (1)

                   [ CH2Cl2]²

={ (x/2) x (x/2) } / { (a-x)² } = 41.1 (given)

Also according to question equilibrium conc. of CCl4 is 0.202 M

i.e. x/2 = 0.202   or x = 0.404 M

Solving the above equation

(a-x) = 0.0315 M

This is the equilibrium conc. of CH2Cl2

To check for correctness put the respective equilibrium conc. values in Eq 1 and check if it returns a value of 41.1

[CH4] = 0.202 M

[CCl4] = 0.202 M

[CH2Cl2] = 0.0315 M

2.

The relationship between Kp and Kc is

Kp = Kc (RT)Δn                                                    Δn is in the power

^Where Δn = No of moles of products - no of moles of reactants

The given reaction is

2HI(g) --> H₂(g) +  I₂(g)

No of moles of gaseous products= 2

No of moles of gaseous reactants =2

Δn = 0

So Kp = Kc, since (RT) to the power zero returns one

Kp = 1.80×10^-2

3

Again Kp = Kc (RT)Δn  .

NH₄Cl(s) --> NH₃(g) +  HCl(g)

Δn = 2-1 = 1

Given temperature = 548 K.

R = 0.0821 L.Atm./K/mole

Kp = 5.10×10^-6 x ( 0.0821x 548) = 229.4x10^-6