Chapter 5, Problem 5.156P

Interpretation Introduction

Interpretation:

The partial pressure of ${\text{I}}_{\text{2}}$ gas is to be calculated.

Concept introduction:

Ideal gas law relates pressure, volume, temperature, and number of moles of gas.

The ideal gas equation can be expressed as follows,

  $PV=nRT$        (1)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

The expression to calculate the partial pressure of the gas is as follows:

  $P_{\text{gas}}=X_{\text{gas}}\cdot P_{\text{total}}$

Here, $P_{\text{gas}}$ is the partial pressure of the gas, $X_{\text{gas}}$ mole fraction of the gas and $P_{\text{total}}$ is the total pressure.

The expression to calculate the mole fraction of the gas is as follows:

  $X_{\text{gas}}=\frac{\text{Moles of gas}}{\text{Total moles}}$

Here, $X_{\text{gas}}$ is the mole fraction of the gas.

Answer to Problem 5.156P

The partial pressure of ${\text{I}}_{\text{2}}$ gas is $25.2\text{ torr}$.

Explanation of Solution

The balanced chemical equation for the reaction of ${\text{I}}_{\text{2}}$ with ${\text{F}}_{\text{2}}$ is as follows:

  ${\text{7F}}_{\text{2}}\left(g\right){\text{ + I}}_{\text{2}}\left(s\right)\to 2{\text{IF}}_7\left(g\right)$        (2)

Convert pressure from $\text{torr}$ to $\text{atm}$ is as follows:

  $\begin{array}{c}P=35\text{0 torr}\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =0.460526315\text{ atm}\end{array}$

The expression to calculate the moles of the ${\text{F}}_{\text{2}}$ is as follows:

  $PV=nRT$        (2)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of ${\text{F}}_{\text{2}}$ and $R$ is the gas constant.

Rearrange equation (1) for $n$.

  $n=\frac{PV}{RT}$        (3)

Substitute $0.460526315\text{ atm}$ for $P$, $250\text{K}$ for $T$, $\text{2}\text{.50 L}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in equation (3) to calculate moles of ${\text{F}}_{\text{2}}$.

  $\begin{array}{c}n=\frac{\left(0.460526315\text{ atm}\right)\left(\text{2}\text{.50 L}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(250\text{ K}\right)}\\ =0.056093339\text{ mol}\end{array}$

The expression to calculate the moles of ${\text{I}}_{\text{2}}$ is as follows:

  ${\text{Moles of I}}_{\text{2}}=\frac{{\text{Mass of I}}_{\text{2}}}{{\text{Molar mass of I}}_{\text{2}}}$        (4)

Substitute $2.50\text{ g}$ for the mass of ${\text{I}}_{\text{2}}$ and $253.8\text{g/mol}$ for molar mass of ${\text{I}}_{\text{2}}$ in the equation (4).

  $\begin{array}{c}{\text{Moles of I}}_{\text{2}}=\frac{2.50\text{ g}}{253.8\text{g/mol}}\\ =0.009850\text{ mol}\end{array}$

According to the balanced chemical equation (2), seven moles of ${\text{F}}_{\text{2}}$ yield two moles of ${\text{IF}}_7$ hence, moles of ${\text{IF}}_7$ produced from ${\text{F}}_{\text{2}}$ is calculated as follows:

  ${\text{Moles of IF}}_{\text{7}}=\left[\left({\text{Moles of F}}_{\text{2}}\right)\left(\frac{{\text{2 mol I}}_{\text{2}}}{{\text{7 mol F}}_{\text{2}}}\right)\right]$        (5)

Substitute $0.056093339\text{ mol}$ for moles of ${\text{F}}_{\text{2}}$ in equation (5).

  $\begin{array}{c}{\text{Moles of IF}}_{\text{7}}=\left[\left(0.056093339\text{ mol}\right)\left(\frac{{\text{2 mol IF}}_{\text{7}}}{{\text{7 mol F}}_{\text{2}}}\right)\right]\\ =0.016026668\text{mol}\end{array}$

From the equation (2), one mole of ${\text{I}}_{\text{2}}$ yields two moles of ${\text{IF}}_7$ hence, moles of ${\text{IF}}_7$ produced from ${\text{I}}_{\text{2}}$ is calculated as follows:

  ${\text{Moles of IF}}_{\text{7}}=\left[\left({\text{Moles of I}}_{\text{2}}\right)\left(\frac{{\text{2 mol IF}}_{\text{7}}}{{\text{1 mol I}}_{\text{2}}}\right)\right]$        (6)

Substitute the value $0.009850\text{ mol}$ for moles of ${\text{I}}_{\text{2}}$ in the equation (6).

  $\begin{array}{c}{\text{Moles of IF}}_{\text{7}}=\left[\left(0.009850\text{ mol}\right)\left(\frac{{\text{2 mol IF}}_{\text{7}}}{{\text{1 mol I}}_{\text{2}}}\right)\right]\\ =0.019700551\text{ mol}\end{array}$

${\text{F}}_{\text{2}}$ is the limiting reagent because it produces less number of moles of ${\text{IF}}_7$.

The stoichiometric ratio between ${\text{I}}_{\text{2}}$ and ${\text{F}}_{\text{2}}$ is 1:7. ${\text{I}}_{\text{2}}$ is excess reagent and thus the amount of ${\text{I}}_{\text{2}}$ reacted is calculated as,

  ${\text{moles of I}}_{\text{2}}{\text{ react with F}}_{\text{2}}=\left({\text{moles of F}}_{\text{2}}\right)\left(\frac{1{\text{ mol I}}_2}{7{\text{ mol F}}_2}\right)$        (7)

Substitute $0.056093339\text{ mol}$ for moles of ${\text{F}}_{\text{2}}$ in equation (7).

  $\begin{array}{c}{\text{moles of I}}_{\text{2}}{\text{ react with F}}_{\text{2}}=\left(0.056093339\text{ mol}\right)\left(\frac{1{\text{ mol I}}_2}{7{\text{ mol F}}_2}\right)\\ =0.00801333414{\text{ mol I}}_2\end{array}$

The expression to calculate the mole ${\text{I}}_{\text{2}}$ left after the reaction is as follows:

  ${\text{Moles of I}}_{\text{2}}=\left[\left({\text{initial moles of I}}_{\text{2}}\right)-\left({\text{moles of I}}_{\text{2}}{\text{ react with F}}_{\text{2}}\right)\right]$        (8)

Substitute the value $0.009850\text{ mol}$ for initial moles of ${\text{I}}_{\text{2}}$ and $0.0080133414\text{mol}$ for moles of ${\text{I}}_{\text{2}}$ reacted with ${\text{F}}_{\text{2}}$ in the equation (8).

  $\begin{array}{c}{\text{Moles of I}}_{\text{2}}=\left[\left(0.009850\text{ mol}\right)-\left(0.0080133414\text{mol}\right)\right]\\ =1.83694\times {10}^{-3}\text{ mol}\end{array}$

The total moles of the gas after the reaction is completed is equal to the sum of moles of ${\text{I}}_{\text{2}}$, ${\text{F}}_{\text{2}}$ and ${\text{IF}}_7$.

  $\begin{array}{c}\text{Total moles}=\left({\text{moles of F}}_{\text{2}}\right)+\left({\text{moles of I}}_{\text{2}}\right)+\left({\text{moles of IF}}_{\text{7}}\right)\\ =\left(0\text{ mol}\right)+\left(1.83694\times {10}^{-3}\text{ mol}\right)+\left(0.019700551\text{ mol}\right)\\ =0.0178636\text{mol}\end{array}$

Rearrange the equation (1) for $P$ as follows:

  $V=\frac{nRT}{P}$        (9)

Substitute $0.0178636\text{mol}$ for $n$, $\text{550 K}$ for $T$, $\text{2}\text{.50 mL}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in equation (9) to calculate the total pressure.

  $\begin{array}{c}P=\frac{\left(0.0178636\text{mol}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(\text{550 K}\right)}{\left(\text{2}\text{.50 L}\right)}\\ =0.322652\text{ atm}\end{array}$

Convert pressure from $\text{atm}$ to $\text{torr}$ as follows:

  $\begin{array}{c}P=0.322652\text{ atm}\left(\frac{760\text{ torr}}{1\text{ atm}}\right)\\ =245.21552\text{ torr}\\ \simeq 245\text{ torr}\end{array}$

The expression to calculate the mole fraction of ${\text{I}}_{\text{2}}$ is as follows:

  $X_{\text{iodine}}=\frac{\text{Moles of iodine}}{\text{total moles}}$        (10)

Substitute $1.83694\times {10}^{-3}$ for moles of iodine and $0.0178636\text{mol}$ for total moles in the equation (10).

  $\begin{array}{c}{X}_{\text{iodine}}=\frac{1.83694\times {10}^{-3}\text{ mol}}{0.0178636\text{mol}}\\ =0.102831\end{array}$

The expression to calculate the partial pressure of ${\text{I}}_{\text{2}}$ is as follows:

  $P_{\text{iodine}}=\left(X_{\text{iodine}}\right)\left(P_{\text{total}}\right)$        (11)

Substitute $0.102831$ for $X_{\text{iodine}}$ and $245.21552\text{torr}$ for $P_{\text{total}}$ in the equation(11).

  $\begin{array}{c}{P}_{\text{iodine}}=\left(0.102831\right)\left(245.21552\text{torr}\right)\\ =25.215869\text{torr}\\ \approx \text{25}\text{.2 torr}\text{.}\end{array}$

Conclusion

The partial pressure of ${\text{I}}_{\text{2}}$ gas is $25.2\text{ torr}$.