Question

When gaseous F2 and solid I2 are heated to high temperatures, the I2 sublimes and gaseous iodine heptafluoride forms.3.30 × 102 torr of F2 and 4.40 g of solid I2 are put into a 2.50 L container at 2.50 × 102 K and the container is heated to 5.50 × 10^2 K.
(a) What is the final pressure?
Ptotal = ____________atm

(b) What is the partial pressure of I2 gas?
PI2 = _________________atm I2

Answer 1

Answer:

a) $P_T=0.447atm=340.02 torr$

b) $P_{I_2}=0.176atm=133.54 torr$

Explanation:

Hello,

At first, we state the undergoing chemical reaction as:

$7F_2(g)+I_2(s)-->2IF_7(g)$

Now, from the beginning we know that there are 4.40 g of solid iodine and the reacting fluorine is at 330 torr, thus, one finds the reacting moles  as follows:

$n_{I_2}=4.40gI_2*\frac{1molI_2}{254gI_2}=0.0173molI_2$

$n_{F_2}=\frac{330 torr*\frac{1atm}{760 torr}*2.50L}{0.082\frac{atm*L}{mol*K} *K} =0.053molF_2$

and identify the limiting reagent based on the theoretical moles of solid iodine that completely react with 0.0530 moles of gaseous fluorine as:

$0.0530molF_2*\frac{1molI_2}{7molF_2}=0.00756molI_2$

Now, as long as there are 0.0173 moles of solid iodine and the reacting amount is smaller, one states that the fluorine is the limiting reagent and there are 0.0173-0.00756= 0.00974 moles of unreacted solid iodine which are subsequently sublimed.

Under these conditions, the stoichiometry allows us to compute the resulting moles of iodine heptafluoride as:

$n_{IF_7}=0.053molF_2*\frac{2molIF_7}{7molF_2}=0.0151molIF_7$

a) Now, to compute the final pressure, one must include both the gaseous moles of iodine heptafluoride and the sublimed iodine by using the ideal gas equation but now at 550K, as follows:

$n_T=0.0151mol+0.00974mol=0.0248mol\\ P_T=\frac{nRT}{V}=\frac{0.0248mol*0.082\frac{atm*L}{mol*K}*550K}{2.50L} =0.447atm=340.02 torr$

b) Finally, for the partial pressure of the sublimed iodine gas, one just takes the unreacted moles which were sublimed due to the heat to compute the pressure as:

$P_{I_2}=\frac{n_{I_2}RT}{V}=\frac{0.00974mol*0.082\frac{atm*L}{mol*K}*550K}{2.50L} =0.176atm=133.54 torr$

Best regards.