Question

1. The equilibrium constant, K_c, for the following reaction is 9.52×10^-2at 350 K.

CH₄(g) + CCl₄(g) = 2 CH₂Cl₂(g)

Calculate the equilibrium concentrations of reactants and product when 0.300 moles of CH₄and 0.300 moles of CCl₄are introduced into a 1.00 L vessel at 350 K.

[ CH4]	=	M
[ CCl4]	=	M
[ CH2Cl2]	=	M

2. 2HI(g) =H₂(g) + I₂(g)

If 1.87 moles of HI, 0.335 moles of H₂, and 0.211 moles of I₂ are at equilibrium in a 15.6 L container at 748 K, the value of the equilibrium constant, K_p, is ?

3. The equilibrium constant, K_p, for the following reaction is 2.01 at 500 K:

PCl₃(g) + Cl₂(g) =PCl₅(g)

Calculate the equilibrium partial pressures of all species when PCl₃ and Cl₂, each at an intitial partial pressure of 0.829 atm, are introduced into an evacuated vessel at 500 K.

PPCl3	=	atm
PCl2	=	atm
PPCl5	=	atm

4. The equilibrium constant, K_p, for the following reaction is 1.80×10^-2 at 698 K:

2HI(g) = H₂(g) + I₂(g)

Calculate the equilibrium partial pressures of all species when HI(g) is introduced into an evacuated flask at a pressure of 1.29 atm at 698 K.

PHI	=	atm
PH2	=	atm
PI2	=	atm

Answer 1

1.

The given equilibrium reaction is

Now, we can write the expression of the equilibrium constant Kc as follows:

Now, given the initial number of moles and volume of the reaction vessel, we can calculate the initial concentrations as follows:

Now, we can create the following ICE chart to calculate the equilibrium concentrations.

Initial, M	0.300	0.300	0
Change, M	-x	-x	+2x
Equilibrium, M	0.300-x	0.300-x	2x

Note that for x moles each of CH₄ and CCl₄ consumed, we get 2x moles of CH₂Cl₂ according to stoichiometry of the reaction.

Hence, we can use the ICE table to write the expression of Kc as follows:

Hence, the equilibrium concentrations can be calculated as follows:

2.

The given equilibrium reaction is

We are given the number of moles of each reactant and product at equilibrium as follows:

For calculating the equilibrium constant Kp, we have to calculate the partial pressures of each gas at T = 748 K and V = 15.6 L.

Hence,

Now, we can calculate the value of K_p as follows:

3.

The given equilibrium is

Hence, the expression of equilibrium constant Kp in terms of partial pressures is

Now, Given that

We can create the following ICE table to calculate the equilibrium partial pressures

Initial, atm	0.829	0.829	0
Change, atm	-x	-x	+x
Equilibrium, atm	0.829-x	0.829-x	x

Now, we can write the expression of Kp using the ICE table as follows:

Hence, the equilibrium partial pressures can be calculated as

4.

The equilibrium reaction is

The expression of equilibrium constant Kp in terms of partial pressure is

Now, given that the initial partial pressure of HI is 1.29 atm, we can create the following ICE table to calculate the equilibrium partial pressure.

Initial, atm	1.29	0	0
Change, atm	-2x	+x	+x
Equilibrium, atm	1.29-2x	x	x

Note that for 2x change in partial pressure of HI, we have +x change in partial pressure of H₂ and I₂.

Now, using the ICE table, we can write the expression of Kp as follows:

Hence, the equilibrium partial pressures can be calculated as

Note: All answers are rounded to three significant figures.