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 H, 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+I2  <——–>    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.