Numericals of Hydrogen Iodide

Hydrogen Iodide

source : ovguide.com

Q.1      In the reacⁿ , the conc. of H₂, I₂ & HI at equi. are  8 , 3 , 28 mole/ litre respectively Determine the equilibrium  constt.

Ans.    [H₂] = 8 mole/l

[I₂]=3 mole/l

[HI] = 28 mole/l

Kc=[HI]2/[H2][I2]

Kc=28 x 28 /8 x 3= 32.67 mole /litre Ans.

Q  2.      In the reacⁿ  

H2 +I2  <——> 2HI , the amount of H₂ , I₂ & HI are 0.2 gm. 9.2525 gm &  44.8 gm respectively At equi. at certain temperature calculate Kc=?

Ans.    w of H₂= 0.2 gm

            m of H₂= 2

w of I₂= 9.2525 gm

m of I₂= 254

w of HI= 44.8

m of HI= 128

[H2] =w/mV = 2/2V= 1/V mole /litre

[I2] = 9.2525/254V =0.36/V mole /litre

[HI] =44.8/128 V= 0.35/V mole/litre

Kc=[HI]2/[H2][I2]

=[(0.35/V) x( 0.35/v)] /[(1/V) (0.36/V)]

Kc =0.34 Ans.

Q3.      At 444⁰C, 16.2 mole of H₂ & 5.88 moles of I₂ and heated. At equi. 11.28 moles of HI is obtained Cal Kc

Ans                             H₂                    +          I₂       <——–>      2HI

Initial Conc.                16.2                             5.88                            O

(mole)

At Equi.                        16.2-5.64                   5.88-5.64                       11.28

= 10.56                       =0.24

2x= 11.28

x= 11.28/2 = 5.64 mole

Kc=[HI]2/[H2][I2]

Kc=([11.28 /V) (11.28/V] /[10.56/V][0.24/V]

Kc = 50.20 Ans.

Q4.      At 444⁰C, 15 mole of H₂, 5.2 moles of I₂ are heated & 10 moles of HI are obtained Kc=?

Ans.    Initial moles of H₂= 15

Initial moles of I₂= 5.2 moles

At Equi. moles of HI=10

                                    H₂                    +          I₂          <——->          2HI

Initial Conc.                  15                                5.2                               O

(mole)

At Equi.                     15-5                            5.2-5.0                        10

= 10.00                       =0.2

2x=10

x= 10/2= 5 moles

Kc=[HI]2/[H2][I2]

Kc=[(10/V)(10/V)]/[10/V][0.2/V]

Kc= 50 Ans.

Q.5  In an experiment 2×127×5.22 gm. I₂ & 2×20.57 gm. H₂ is mixed. At equi. 10.22×128 gm. HI are obtained Kc=?

Ans.    w of H₂= 2×20.57 gm

m of H₂= 2

moles of  H₂= w/m =2 x 20.57/2 =20.57 

w of I₂= 2×127×5.22 gm

m of I₂= 254

moles of  I₂=  w/m=2 x 127 x 5.22/254=5.22 

At equi.,

w of HI= 10.22×128 gm

m of HI= 128

moles of  HI=  w/m =10.22 x 128 /128 =10.22

                                        H₂               +          I₂        <——–>             2HI

Initial Conc. (mole)           20.57               5.22                            O

At Equi.                              20.57-5.11               5.22-5.11                   10.22

= 15.46                       =0.11

Kc=[HI]2/[H2][I2]

Kc=[(10.22/V)(10.22/V)]/[15.46/V][0.11/V]

Kc   =61.42 Ans.

Q.6     0.492 gm H₂ & 31.5 gm. I₂ are heated in a closed vessel of 2l capacity. At 460C at equi. 29.57 gm of HI is obtained. calculate the value of equi. constt. for the reaction    H₂+I_{2  <——–>}    2HI

Ans.    w of H₂ = 0.492 gm

m of H₂ = 2

initial moles of H₂   =w/m =0.492/2 = 0.246

            w of I₂ = 31.5 gm

m of I₂ = 254

initial moles of I₂ =w/m= 31.5 /254 = 0.124

At equi.,

w of HI = 29.57 gm

m of HI = 128

moles of HI  =w/m =29.57/128 = 0.23

                                        H₂               +          I₂          <——–>           2HI

Initial Conc.                   0.246                         0.124                          O

At Equi.                       0.246-0.115           0.124-0.115               0.231

= 0.131                       =0.009

2x= 0.231

x= 0.231/2 = 0.115

Kc=[HI]2/[H2][I2]

Kc=[(0.231/V)(0.231/V)] / [0.131/V][0 .009 /V]

Kc = 45.26 Ans.

Q.7      At 444⁰C in a vessel of 1 litre , 2.85 moles of Hydrogen iodide is heated in presence of 1 mole of H2 till the equi. is attained. At equilibrium 0.1 mole of I₂ is obtained. calculate the value of equilibrium Constt.

Ans.                               2HI     <——->   H₂                    +          I₂

Initial Conc.                   2.85/V                    0                                 0

At Equi.                          2.85 -0.20 /V      (  1 +  0.1) /V            0.1 /V

-2.65 /V                   =1.10/V                      0.10/V

Kc  =[H2][I2]/[HI]2

=[ (1.10/V)(0.10/V)] / [2.65/V]2

Kc= 0.156 Ans.

Q.8 At 444⁰C, 20 ml H₂ & 20 ml of I₂ are heated at equi., 32 ml of Hydrogen iodide is obtained. Calculate Kc?

Ans                                 H₂               +          I₂         <——->            2HI

Initial Vol.                       20                            20                            0

Vol. At Equi.                   20-16             20-16                          32

= 4                        =4

2x=32

x= 32/2=16

Kc=[HI]2/[H2][I2]

Kc=[32]2/[4][4]

Kc= 64 Ans.

Note At constt. temp. & Pressure, volume of a gas is proportional to no. of moles of gas. Hence volume is used in place of molar concentration in equilibrium expression.

Q9.      At const. temp. 1.2 mole of HI is heated. At equi. 0.12 mole of I2 is obtained. Calculate the equi. constt. for the reacⁿ

Ans.                            2HI              <——>                    H₂                    +          I₂

Initial Conc.                  1.2/V                                     0                                  0

Concⁿ at Equi.              1.2 -0.24/V                        0 .12/V                         0 .12/V

Kc =[H2][I2] /[HI]2

=[ (0.12/V) (0.12/V)]/[0.96/V]2

Kc= 0.0156 Ans.

           Q.10   At 440⁰C, 2moles of HI are heated in closed  vessel at equi, 22% of HI is dissociated calculated the Kc for the reacⁿ

Amount of Hydrogen iodide Dissociated =2 x 22/100 =0.44 mole

                                    2HI                <—->                  H₂                    +          I₂

             ^{initial conc.                2 /V                                                    0                                            0}

conc. at equi.       =1.56/V                             0.22/V                        0.22/V

Kc =[H2][I2] /[HI]2

Kc =[0.22/V][0.22/V] /[1.56/V]2

Kc= 0.0198 Ans.