Question

When gaseous F2 and solid I2 are heated to high temperatures, the I2 sublimes and gaseous iodine heptafluoride forms. If 350. torr of F2 and 2.50 g of solid I2 are put into a 2.50-L container at 250. K and the container is heated to 550. K, what is the final pressure (in torr)? What is the partial pressure of I2 gas?

Answer 1

Answer:

$P_{I2}=0.033atm$

Explanation:

Hi, the first step is to calculate how much F2 there is in the container:

Fluorine can be considered as an ideal gas (given that is non-polar and has a small molecule). Using the ideal gas formula:

$n=\frac{V*P}{T*R}$

Where:

$P=350 torr * \frac{1 atm}{760 torr}=0.46atm$

$T=250K$

$V=2.5 L$

$R=0.082 \frac{atm*L}{mol*K}$

$n=\frac{2.5L*0.46atm}{250K*0.082 \frac{atm*L}{mol*K}}$

$n=0.056 mol$

Now, the mols of iodine:

$n_{I2}=\frac{2.5g}{253.8 g/mol}$

$n_{I2}=9.85*10^{-3}mol$

The chemical reaction described is the following:

$I_2(g) + 7 F_2(g) \longrightarrow 2 IF_7(g)$

In this case, the limitant reactant is the fluorine:

1) The 0.056 mol of F2 gives $8*10^{-3} mol$ of $IF_7$ and consumes $8*10^{-3} mol$ of I2.

2) At the end, in the conteiner we have:

$8*10^{-3} mol$ of $IF_7$

$0 mol$ of $F_2$

$9.85*10^{-3}-8*10^{-3}=1.85*10^{-3} mol$ of $I_2$

In total: $9.85*10^{-3}mol$ .All in 2.5 L at 550 K

The final pressure:

$P=\frac{T*R*n}{V}$

$P=\frac{550K*0.082 \frac{atm*L}{mol*K}*9.85*10^{-3}mol}{2.5L}$

$P=0.176 atm$

The partial pressure:

$P_{I2}=0.176atm*\frac{1.85*10^{-3} mol (I2)}{9.85*10^{-3} mol (total)}$

$P_{I2}=0.033atm$

Note: this partial pressure is calculated by the Dlaton's principle