3.2 moles of HI(g) is heated in a sealed bulb at 444°C till the equilibrium was reached. It's degree of dissociation was found to be 20%.
calculate the number of moles of hydrogen iodide, hydrogen and iodine present at the equilibrium point and determine the value of equilibrium constant for the reaction 2HI(g) H2(g)+I2(g). Considering the volume of the container 1L
calculate the number of moles of hydrogen iodide, hydrogen and iodine present at the equilibrium point and determine the value of equilibrium constant for the reaction 2HI(g) H2(g)+I2(g). Considering the volume of the container 1L
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=3.2M
⭕️degree of dissociation= 20%
⭕️degree of dissociation = no of moles of reactants dissociated/no. of moles of reactants present initially
20/100=x/3.2
x=0.64 moles
2HI(g)=H2(g)+I2(g)
⭕️two M of HI gives one M of H2 and I2 each so....
64 M of HI would give 32 M of H2 and I2 each
⭕️hence equilibrium concentration of HI is 2.56M
⭕️2.56 is obtained by 3.2 - 0.64 which is number of moles left at equilibrium.
Kc=[0.32][0.32]/[2.56]2
=0.0156
===== Answer ======
Concentration of HI = 3.2 Moles /litre or 3.2 M
Degree of dissociate = 20%
= no of Moles of reactant dissociated / no of Moles of reactant initially
= 20/100 = X/3.2
So...
2 HI (g) = H2(g) + 12 g
2 Moles of HI gives 1 mole of H2 & 12 each so...
64 Moles of HI give 32 Moles of H2
Equilibrium concentration is = 2.56
KC = [0.32][0.32]/ [2.56]2 = 0.0156
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