Hydrogen Iodide
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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+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.
