Question

1 mole of hydrogen and 2 moles of iodine are taken initially in a 2 litre vessel. the no: of moles of hydrogen at equilibrium is 0.2. then the no: of moles of iodine and hydrogen iodide at equilibrium are:

Answer 1

H2 + I2 ---> 2HI
at initial
1/2 mole/l of H2
2/2 mole/l of I2
and zero mole/l

now at equilibrium
(0.5-x)mole/l of H2
(1-x) mole/l of I2
and (1-2x)mole/l
of HI
here we see limiting reagents is H2 hence
concentration of HI is related to H2 not I2

now a/c
0.5-x=0.2/2=0.1
x=0.4mole/l
hence I2 concentration is 1-0.4=0.6mole/l
hence no of mole of I2
=0.6x2=1.2mole

now concentration of HI=1-2(0.1)=0.8
because concentration of H2=0.1mole/l
now no of mole of HI=0.8x2=1.6

Answer 2

Answer:

Explanation:

1.2  and 1.6

Solution :

                              H2(g)    +   I2(g)     ⇔2HI(g)  

Initial moles            1                 2             0

Equilibrium moles  1−x            2-x       ⇔2x

nH2  = 1−x        = 0.2

or x   =  1.0−0.2 =  0.8

∴nI2  =2−x         =2−0.8   =1.2

nHI    =2x           =2(0.8)   =1.6

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