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

`P_(I2)=0.033atm`

Step-by-step 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=(V*P)/(T*R)`

Where:

`P=350 torr * (1 atm)/(760 torr)=0.46atm`

`T=250K`

`V=2.5 L`

`R=0.082 (atm*L)/(mol*K)`

`n=(2.5L*0.46atm)/(250K*0.082 (atm*L)/(mol*K))`

`n=0.056 mol`

Now, the mols of iodine:

`n_(I2)=(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=(T*R*n)/(V)`

`P=(550K*0.082 (atm*L)/(mol*K)*9.85*10^(-3)mol)/(2.5L)`

`P=0.176 atm`

The partial pressure:

`P_(I2)=0.176atm*(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