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damaskus [11]
3 years ago
15

When gaseous F2 and solid I2 are heated to high temperatures, the I2 sublimes and gaseous iodine heptafluoride forms. If 350. to

rr 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?
Chemistry
1 answer:
lilavasa [31]3 years ago
8 0

Answer:

P_{I2}=0.033atm

Explanation:

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

<u>Fluorine can be considered as an ideal gas</u> (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

<u>The chemical reaction described is the following</u>:

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

<u>The final pressure</u>:

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

<em>Note: this partial pressure is calculated by the Dlaton's principle</em>

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