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Question: 3.2 moles of HI (g) were heated in a sealed bulb at \({\text{44}}{{\text{4}}^{\text{o}}}{\text{C}}\)...

3.2 moles of HI (g) were heated in a sealed bulb at 444oC{\text{44}}{{\text{4}}^{\text{o}}}{\text{C}} till the equilibrium was reached. Its 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){\text{2H}}{{\text{I}}_{\left( {\text{g}} \right)}} \rightleftharpoons {{\text{H}}_{\text{2}}}_{\left( {\text{g}} \right)}{\text{ + }}{{\text{I}}_{\text{2}}}_{\left( {\text{g}} \right)}.Considering the volume of the container 1L.

Explanation

Solution

The number of moles of all the species in the above reaction can be known from the degree of dissociation (α\alpha) and from the initial moles of HI. Degree of dissociation is the extent of which the given molecules of the substance are dissociated or separated out.

Complete Solution :
In the question it is given that, initially 3.2 moles of HI was heated in a sealed tube to about a temperature of 444oC{\text{44}}{{\text{4}}^{\text{o}}}{\text{C}}, till it reaches the equilibrium and the degree of dissociation is given as 20% or it can be also taken as 0.2 as per the percentage calculation. And we have to find the equilibrium constant and number of moles of HI, H and I at equilibrium.

The given reaction is :2HI(g)⇌H2(g) + I2(g){\text{2H}}{{\text{I}}_{\left( {\text{g}} \right)}} \rightleftharpoons {{\text{H}}_{\text{2}}}_{\left( {\text{g}} \right)}{\text{ + }}{{\text{I}}_{\text{2}}}_{\left( {\text{g}} \right)}

At time, t=o| 3.2 moles| 0 moles| 0 moles
After some time,t=t sec| 3.2-(2x) moles| x moles| x moles

Moles dissociated is = 3.2×20100\dfrac{3.2\times 20}{100}
Moles dissociated is = 0.64 moles = (2x)
So to find out the concentrations of all the species of the above reaction let us substitute the value of 2x at time t.

And the number of moles of HI, H and I at equilibrium is,
The number of moles of HI is, 3.2-(2x) = 3.2-0.64 = 2.56 moles
The number of moles of hydrogen is x, 0.64/2 = 0.32moles
The number of moles of iodine is x, 0.64/2 = 0.32moles.

And now we have to find the equilibrium constant of the reaction (Kc{{K}_{c}}), which can be calculated by the formulae,
Kc=[H2][I2][HI]2{{K}_{c}}=\dfrac{\left[ {{H}_{2}} \right]\left[ {{I}_{2}} \right]}{{{\left[ HI \right]}^{2}}} Kc{{K}_{c}}
Kc=[0.32][0.32][2.56]2{{K}_{c}}=\dfrac{\left[ 0.32 \right]\left[ 0.32 \right]}{{{\left[ 2.56 \right]}^{2}}}
Kc{{K}_{c}} = 0.0156

Note: Since the degree of dissociation given is 20%, it means that 20% of 3.2 moles of HI has been dissociated and the final concentration remaining at equilibrium would be the 0.2 times that of 3.2 moles.