Chapter 6, Problem 79AE

a)

Interpretation Introduction

Interpretation:Partial pressure of each gasin below reaction is to be determined.

  ${\text{PCl}}_{\text{5}}\left(g\right)\rightleftharpoons {\text{PCl}}_{\text{3}}\left(g\right)+{\text{Cl}}_2\left(g\right)$

Concept introduction: Chemical equilibrium is taken into consideration if rate of forward and backward reactions becomes equal. At this stage, both reactants and products have constant concentration. It can be studied in terms of pressure also. Equilibrium constant in pressure is denoted by $K_{\text{p}}$ .

Explanation of Solution

Given Information: Mass of ${\text{PCl}}_5$ is $2.4156\text{ g}$ at $250.0°\text{C}$ . At equilibrium, total pressure of mixtureis $358.7\text{ torr}$ .

Given reaction occurs as follows:

  ${\text{PCl}}_{\text{5}}\left(g\right)\rightleftharpoons {\text{PCl}}_{\text{3}}\left(g\right)+{\text{Cl}}_2\left(g\right)$

Moles of ${\text{PCl}}_5$ can be calculated as follows:

  $\begin{array}{c}\text{Moles}=\left(\frac{2.4156\text{ g}}{208.24\text{ g/mol}}\right)\\ =0.01160\text{ mol}\end{array}$

Conversion of $250.0°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =250.0°\text{C}+273\\ =523\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

  $P$ is pressure of the gas.

  $V$ is volume of gas.

  $n$ denotes moles of gas.

  $R$ is gas constant.

  $T$ is temperature of gas.

Rearrange above equation to calculate value of $P$ for ${\text{Cl}}_{\text{2}}$ .

  $P=\frac{nRT}{V}$

Value of $n$ is $0.250\text{ mol}$ .

Value of $T$ is $523\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $2.000\text{L}$ .

Substitute the value in above equation.

  $\begin{array}{c}P=\frac{nRT}{V}\\ =\frac{\left(0.01160\text{ mol}\right)\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(523\text{K}\right)}{\left(2.000\text{L}\right)}\\ =\left(0.2489\text{ atm}\right)\left(\frac{760\text{ torr}}{1\text{ atm}}\right)\\ =189.2\text{ torr}\end{array}$

TheICE table for the above reaction can be drawn as follows:

  $\begin{array}{cccccc}\text{Equation}& {\text{PCl}}_{\text{5}}& \leftrightarrow & {\text{PCl}}_{\text{3}}& +& {\text{Cl}}_2\\ \text{Initial}\left(\text{torr}\right)& 189.2& & 0& & 0\\ \text{Change}\left(\text{torr}\right)& -x& & +x& & +x\\ \text{Equilibrium}\left(\text{torr}\right)& 189.2-x& & x& & x\end{array}$

Total equilibrium pressure is $358.7\text{ torr}$ . Therefore value of $x$ is calculated as follows:

  $\begin{array}{c}189.2-x+x+x=358.7\text{ torr}\\ x=169.5\text{ torr}\end{array}$

Equilibrium pressure of ${\text{PCl}}_{\text{5}}$ is calculated as follows:

  $\begin{array}{c}{P}_{{\text{PCl}}_{\text{5}}}=\left(189.2-169.5\right)\text{ torr}\\ =19.7\text{ torr}\end{array}$

Equilibrium pressure of ${\text{PCl}}_{\text{3}}$ is calculated as follows:

  $P_{{\text{PCl}}_{\text{3}}}=169.5\text{ torr}$

Equilibrium pressure of ${\text{Cl}}_2$ is calculated as follows:

  $P_{{\text{Cl}}_2}=169.5\text{ torr}$

Expression for $K_{\text{p}}$ of given reaction is as follows:

  $K_{\text{p}}=\frac{\left(P_{{\text{Cl}}_{\text{2}}}\right)\left(P_{{\text{PCl}}_{\text{3}}}\right)}{\left(P_{{\text{PCl}}_{\text{5}}}\right)}$

Where,

• $K_{\text{p}}$ is equilibrium constant.
• $P_{{\text{Cl}}_{\text{2}}}$ is equilibrium partial pressure of ${\text{Cl}}_{\text{2}}$ .
• $P_{{\text{PCl}}_{\text{3}}}$ is equilibrium partial pressure of ${\text{PCl}}_{\text{3}}$ .
• $P_{{\text{PCl}}_{\text{5}}}$ is equilibrium partial pressure of ${\text{PCl}}_{\text{5}}$ .

Value of $P_{{\text{Cl}}_{\text{2}}}$ is $169.5\text{ torr}$ .

Value of $P_{{\text{PCl}}_{\text{3}}}$ is $169.5\text{ torr}$ .

Value of $P_{{\text{PCl}}_{\text{5}}}$ is $19.7\text{ torr}$ .

Substitute the values in above equation.

  $\begin{array}{c}{K}_{\text{p}}=\frac{\left(P_{{\text{Cl}}_{\text{2}}}\right)\left(P_{{\text{PCl}}_{\text{3}}}\right)}{\left(P_{{\text{PCl}}_{\text{5}}}\right)}\\ =\frac{\left(169.5\text{ torr}\right)\left(169.5\text{ torr}\right)}{\left(19.7\text{ torr}\right)}\\ =1458\end{array}$

Hence, value of $K_{\text{p}}$ is 1458.

b)

Interpretation Introduction

Interpretation:New equilibrium pressure of each gasif $0.250{\text{ mol Cl}}_2$ is added in below reaction is to be determined.

  ${\text{PCl}}_{\text{5}}\left(g\right)\rightleftharpoons {\text{PCl}}_{\text{3}}\left(g\right)+{\text{Cl}}_2\left(g\right)$

Concept introduction: Chemical equilibrium is taken into consideration if rate of forward and backward reactions becomes equal. At this stage, both reactants and products have constant concentration. It can be studied in terms of pressure also. Equilibrium constant in pressure is denoted by $K_{\text{p}}$ .

Explanation of Solution

Given Information: Mass of ${\text{PCl}}_5$ is $2.4156\text{ g}$ at $250.0°\text{C}$ . At equilibrium, total pressure of mixture is $358.7\text{ torr}$ . Moles of ${\text{Cl}}_2$ added to reaction mixture is $0.250\text{ mol}$ .

Given reaction occurs as follows:

  ${\text{PCl}}_{\text{5}}\left(g\right)\rightleftharpoons {\text{PCl}}_{\text{3}}\left(g\right)+{\text{Cl}}_2\left(g\right)$

Concentration of ${\text{Cl}}_2$ can be calculated as follows:

  $\begin{array}{c}\text{Moles}=\left(\frac{0.250\text{ mol}}{2.000\text{ L}}\right)\\ =0.125M\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of the gas.
• $V$ is volume of gas.
• $n$ denotes moles of gas.
• $R$ is gas constant.
• $T$ is temperature of gas.

Rearrange above equation to calculate value of $P$ for ${\text{Cl}}_{\text{2}}$ .

  $P=\frac{nRT}{V}$

Value of $n$ is $0.250\text{ mol}$ .

Value of $T$ is $523\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $2.000\text{L}$ .

Substitute the value in above equation.

  $\begin{array}{c}P=\frac{nRT}{V}\\ =\frac{\left(0.250\text{ mol}\right)\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(523\text{K}\right)}{\left(2.000\text{L}\right)}\\ =\left(5.36\text{ atm}\right)\left(\frac{760\text{ torr}}{1\text{ atm}}\right)\\ =4070\text{ torr}\end{array}$

The ICE table for the above reaction can be drawn as follows:

  $\begin{array}{cccccc}\text{Equation}& {\text{PCl}}_{\text{5}}& \rightleftharpoons & {\text{PCl}}_{\text{3}}& +& {\text{Cl}}_2\\ \text{Initial}\left(\text{torr}\right)& 19.7& & 169.5& & 169.5+4070\\ \text{Change}\left(\text{torr}\right)& +x& & -x& & -x\\ \text{Equilibrium}\left(\text{torr}\right)& 19.7+x& & 169.5-x& & 4239.5-x\end{array}$

Expression for $K_{\text{p}}$ of given reaction is as follows:

  $K_{\text{p}}=\frac{\left(P_{{\text{Cl}}_{\text{2}}}\right)\left(P_{{\text{PCl}}_{\text{3}}}\right)}{\left(P_{{\text{PCl}}_{\text{5}}}\right)}$

Where,

• $K_{\text{p}}$ is equilibrium constant.
• $P_{{\text{Cl}}_{\text{2}}}$ is equilibrium partial pressure of ${\text{Cl}}_{\text{2}}$ .
• $P_{{\text{PCl}}_{\text{3}}}$ is equilibrium partial pressure of ${\text{PCl}}_{\text{3}}$ .
• $P_{{\text{PCl}}_{\text{5}}}$ is equilibrium partial pressure of ${\text{PCl}}_{\text{5}}$ .

Value of $P_{{\text{Cl}}_{\text{2}}}$ is $4239.5-x$ .

Value of $P_{{\text{PCl}}_{\text{3}}}$ is $169.5-x$ .

Value of $P_{{\text{PCl}}_{\text{5}}}$ is $19.7+x$ .

Value of $K_{\text{p}}$ is 1458.

Substitute the values in above equation.

  $\begin{array}{c}{K}_{\text{p}}=\frac{\left(P_{{\text{Cl}}_{\text{2}}}\right)\left(P_{{\text{PCl}}_{\text{3}}}\right)}{\left(P_{{\text{PCl}}_{\text{5}}}\right)}\\ 1458=\frac{\left(4239.5-x\right)\left(169.5-x\right)}{\left(19.7+x\right)}\end{array}$

Value of $x$ is calculated as follows:

  $\begin{array}{c}\frac{\left(4239.5-x\right)\left(169.5-x\right)}{\left(19.7+x\right)}=1458\\ x=120.0\text{ torr}\end{array}$

New equilibrium pressure $P_{{\text{Cl}}_{\text{2}}}$ can be calculated as follows:

  $\begin{array}{c}{P}_{{\text{Cl}}_{\text{2}}}=4239.5-120.0\text{ torr}\\ =4120\text{ torr}\end{array}$

New equilibrium pressure $P_{{\text{PCl}}_{\text{3}}}$ can be calculated as follows:

  $\begin{array}{c}{P}_{{\text{PCl}}_{\text{3}}}=169.5-120.0\text{ torr}\\ =4120\text{ torr}\end{array}$

New equilibrium pressure $P_{{\text{PCl}}_{\text{5}}}$ can be calculated as follows:

  $\begin{array}{c}{P}_{{\text{PCl}}_{\text{5}}}=19.7+120.0\text{ torr}\\ =1.40\times {10}^2\text{ torr}\end{array}$