Question

Question 1: Consider the following reaction:

SO2Cl2(g)⇌SO2(g)+Cl2(g)
Kc=2.99×10−7 at 227 ∘C

If a reaction mixture initially contains 0.195 MSO2Cl2, what is the equilibrium concentration of Cl2at 227 ∘C??

Question 2:

Consider the reaction

CO(g)+H2O(g)⇌CO2(g)+H2(g)
Kc=102 at 500 K

A reaction mixture initially contains 0.130 MCOand 0.130 MH2O.

What will be the equilibrium concentration of H2O?

Question 3:

The following reaction was performed in a sealed vessel at 768 ∘C :

H2(g)+I2(g)⇌2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.65M and [I2]=2.70M. The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Please explain and answer all! Thank you

Answer 1

Question 1: Consider the following reaction:

SO2Cl2(g)⇌SO2(g)+Cl2(g)
Kc=2.99×10−7 at 227 ∘C

If a reaction mixture initially contains 0.195 MSO2Cl2, what is the equilibrium concentration of Cl2at 227 ∘C??

Answer :-

Given reaction is in equilibrium,

so we can write

	SOCl2	SO2	Cl2
Initial Concentration	0.195	0	0
Change in concentration	-x	+x	+x
Concentration After Equilibrium	0.195 -x	+x	+x

We can write an equation of reaction concentration as

K = [SO2][O2]/[SO2Cl2]

Lets fill the values from table in the formula

2.99 x 10^-7 = (x)(x)/(0.195 - x )

Neglect x in the 0.195 -x ( because X is very small, so lets consider it as only 0.195)

2.99 x 10^-7 = (x)(x)/0.195

x^2 = 5.8305*10^-8

x = 2.4146 *10^-4

X is nothing but the concentration of Cl2 gas

So, x = 2.18 x 10-4 M is our final answer.

Consider the reaction

CO(g)+H2O(g)⇌CO2(g)+H2(g)
Kc=102 at 500 K

A reaction mixture initially contains 0.130 MCOand 0.130 MH2O.

What will be the equilibrium concentration of H2O?

Answer :-

Given reaction is in equilibrium,

so we can write

	CO(g)	H2O(g)	CO2(g)	H2(g)
Initial Concentration	0.130 M	0.130 M	0	0
Change in concentration	-x	-x	+x	+x
Concentration After Equilibrium	0.130 -x	0.130 -x	+x	+x

We can write an equation of reaction concentration as

Ka = [CO2] [H2] / ([CO] [H2O])

Lets put the values in the formula

102 = (X)(X)/ (0.130 –x)*( 0.130 –x)

102 = (X)^2 / ( 0.130 –x)^2

Lest take a square root of this equation

10.10 = (X) / ( 0.130 –x)

10.10*(0.130 –x) = (X)

1.3130- 10.10 X= X

1.3130 = X + 10.10 X

1.3130 = 11.10 X

X = (1.3130)/ 11.10

X= 0.1183 M

Equilibrium concentration of H2O = 0.130- X

= 0.130-0.1183 M

= 0.0117 M

This is our final answer

The following reaction was performed in a sealed vessel at 768 ∘C :

H2(g)+I2(g)⇌2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.65M and [I2]=2.70M. The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Answer :-

Given reaction is in equilibrium,

so we can write

H2(g)+I2(g)⇌2HI(g)

As per the given reaction

When 1 mole of I2 reacts with 1 mole of H2 then 2 moles of product (HI) is formed

So when 0.08 M of I2 reacts with 0.08 M of H2 then 1.6M (2*0.08M) of HI forms

	H2	I2	2HI
Initial Concentration	3.65	2.70	0
Change in concentration	0.0800	0.0800(given )	1.6
Concentration After Equilibrium	(3.65-0.0800) =3.57	(2.70-0.0800)= 2.62	1.6

We can write an equation of reaction concentration as

K = [HI]^2/[H2] [I2]

Lets fill the values from table in the formula

K = [1.6]^2/[3.57] [2.62]

lets solve this for K

K = [1.6]^2/[3.57] [2.62]

K = [2.56]/[9.3534]

K =0.27370