3.2 moles of HI(g) were 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. plz answer me soon and explain
Answers
Answered by
346
Concentration of HI is 3.2moles/1litrre=3.2
Degree of dissociation=20%
Degree of dissociation=no.of moles of reactant dissociated/no.of moles of reactants present initially 20/100=x3/2
x=0.64moles
According to the reaction
2Hl(g)=H2(g)+l2(g)
Two M of Hl gives one M of H2 & l2 each so 64M of HI would give 32M OF H2 & I2 each hence equilibrium concentration of HI is 2.56M
Kc=(0.32)(0.32)/(2.56)2=0.0156
Degree of dissociation=20%
Degree of dissociation=no.of moles of reactant dissociated/no.of moles of reactants present initially 20/100=x3/2
x=0.64moles
According to the reaction
2Hl(g)=H2(g)+l2(g)
Two M of Hl gives one M of H2 & l2 each so 64M of HI would give 32M OF H2 & I2 each hence equilibrium concentration of HI is 2.56M
Kc=(0.32)(0.32)/(2.56)2=0.0156
Answered by
33
Answer is in the attachment.
Please Mark it as brainliest...
Attachments:
Similar questions