Answer:
[tex]P_{I2}=0.033atm[/tex]
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:
[tex]n=\frac{V*P}{T*R}[/tex]
Where:
[tex]P=350 torr * \frac{1 atm}{760 torr}=0.46atm[/tex]
[tex]T=250K[/tex]
[tex]V=2.5 L[/tex]
[tex]R=0.082 \frac{atm*L}{mol*K}[/tex]
[tex]n=\frac{2.5L*0.46atm}{250K*0.082 \frac{atm*L}{mol*K}}[/tex]
[tex]n=0.056 mol[/tex]
Now, the mols of iodine:
[tex]n_{I2}=\frac{2.5g}{253.8 g/mol}[/tex]
[tex]n_{I2}=9.85*10^{-3}mol[/tex]
The chemical reaction described is the following:
[tex]I_2(g) + 7 F_2(g) \longrightarrow 2 IF_7(g)[/tex]
In this case, the limitant reactant is the fluorine:
1) The 0.056 mol of F2 gives [tex]8*10^{-3} mol[/tex] of [tex]IF_7[/tex] and consumes [tex]8*10^{-3} mol[/tex] of I2.
2) At the end, in the conteiner we have:
[tex]8*10^{-3} mol[/tex] of [tex]IF_7[/tex]
[tex]0 mol[/tex] of [tex]F_2[/tex]
[tex]9.85*10^{-3}-8*10^{-3}=1.85*10^{-3} mol[/tex] of [tex]I_2[/tex]
In total: [tex]9.85*10^{-3}mol[/tex] .All in 2.5 L at 550 K
The final pressure:
[tex]P=\frac{T*R*n}{V}[/tex]
[tex]P=\frac{550K*0.082 \frac{atm*L}{mol*K}*9.85*10^{-3}mol}{2.5L}[/tex]
[tex]P=0.176 atm[/tex]
The partial pressure:
[tex]P_{I2}=0.176atm*\frac{1.85*10^{-3} mol (I2)}{9.85*10^{-3} mol (total)}[/tex]
[tex]P_{I2}=0.033atm[/tex]
Note: this partial pressure is calculated by the Dlaton's principle
