Question

When gaseous F₂ and solid I₂ are heated to high temperatures, the I₂ sublimes and gaseous iodine heptafluoride forms. 280 torr of F₂ and 4.00g of I₂ are put into a 2.50L container at 250K and the container is heated to 550K.

a) what is the final pressure in atm?

b) what is the partial pressure of I₂ gas in atm?

Answer 1

I₂ + 7F₂$\rightarrow$ 2IF₇

No of moles of I₂ = given mass / molar mass = 4 g / 253.8 g mol^-1 = 0.0158 mol

No of moles of F₂ :  

We know, PV = nRT   $\Rightarrow$ n = PV/RT

P_F2 = 280 torr = 0.368 atm, V = 2.5 L, T = 250 K, R = 0.082 L atm K^-1 mol^-1

n = ( 0.368 atm X 2.5 L) / (0.082 L atm K^-1 mol^-1 X 250 K) = (0.92 / 20.5) mol = 0.0449 mol

1 mol of I₂ reacts with 7 moles of F₂     $\Rightarrow$ 0.0158 mol of I₂ will completely react with 7 X 0.0158 = 0.1106 mol of F₂  

But this is lesser than the number of moles of  F₂ available. It means that I₂ is present in excess and that  F₂ is the limiting reagent.

7 mol of F₂ reacts with 1 moles of I₂     $\Rightarrow$ 0.0449 mol of F₂ will completely react with (1/7) X 0.0449 = 0.0064 mol of I₂

Remaining moles of I₂ = 0.0158 mol - 0.0064 mol = 0.0094 mol

Also, 7 mol of F₂ form 2 moles of IF₇     $\Rightarrow$ 0.0449 mol of F₂ will form (2/7) X 0.0449 = 0.0128 mol = No of moles of IF₇ formed.

After the reaction is complete there are 0.0128 mol IF₇ + 0.0094 mol I₂

So, total no of moles after completion of rxn = 0.0222 mol

(a) P_final = (n_Total RT) / V      $\Rightarrow$ P_final = (0.0222 mol X 0.082 L atm K^-1 mol^-1 X 550 K) / 2.5 L

P_final = 0.4005 atm

(b) p_I2 = partial pressure of I₂ , P_T = Total pressure

x_I2 = mole fraction of I₂ = moles of I₂ / total no of moles = 0.0094 mol / 0.0222 mol = 0.4234

p_I2 = x_I2 X P_T = 0.4234 X 0.4005 atm = 0.1696 atm

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