Question

Given K = 3.50 at 45^oC for the reaction

$\mathbf{A}\left(g\right)\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{B}\left(g\right)\mathbf{\rightleftharpoons }\mathbf{C}\left(g\right)$

and K = 7.10 at 45^oC for the reaction

$\mathbf{2}\mathbf{A}\left(g\right)\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{D}\left(g\right)\mathbf{\rightleftharpoons }\mathbf{C}\left(g\right)$

what is the value of K at the same temperature for the reaction

$\mathbf{C}\left(g\right)\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{D}\left(g\right)\mathbf{\rightleftharpoons }\mathbf{2}\mathbf{B}\left(g\right)$

What is the value of K_p at 45^oC for the reaction? Starting with 1.50 atm partial pressures of both C and D, what is the mole fraction of B once equilibrium is reached?

Short answer

The value of K_p at 45^oC for the reaction$\mathrm{C}\left(\mathrm{g}\right) + \mathrm{D}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{B}\left(\mathrm{g}\right)$is found to be 0.579.

The mole fraction of B once equilibrium reached is 0.275.

Step-by-step solution

Calculate the equilibrium constant values.

The equilibrium constants for the below two reactions are named as follows:

$$\begin{gathered} 2\mathrm{A}\left(\mathrm{g}\right) + 2\mathrm{B}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{C}\left(\mathrm{g}\right){\mathrm{K}}_1 \\ 2\mathrm{A}\left(\mathrm{g}\right) + \mathrm{D}\left(\mathrm{g}\right)\rightleftarrows \mathrm{C}\left(\mathrm{g}\right){\mathrm{K}}_2 \end{gathered}$$

The equilibrium constant for the reaction$\mathrm{C}\left(\mathrm{g}\right) + \mathrm{D}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{B}\left(\mathrm{g}\right)$ can be calculated as follows:

$$\begin{gathered} 2C\left(g\right)\rightleftharpoons 2A\left(g\right) + 2B\left(g\right)1/{K_1}^2 \\ \underline{2A\left(g\right) + D\left(g\right)\rightleftharpoons C\left(g\right)K_2} \\ C\left(g\right) + D\left(g\right)\rightleftharpoons 2B\left(g\right)K=\frac{K_2}{{K_1}^2} \end{gathered}$$

Substitute the values of equilibrium constants K₁ and K₂ into the above expression as follows:

$\begin{array}{c}\mathrm{K}=\frac{7.10}{{\left(3.50\right)}^2}\\ =0.579\end{array}$

Therefore, the value of K at the same temperature for this reaction is found to be 0.579.

Determine the relationship between Kp and K for the given reaction.

The below expression that describes the relationship between K_p and K is,

${\mathrm{K}}_{\mathrm{p}}=\mathrm{K}{\left(\mathrm{RT}\right)}^{\mathrm{\Delta n}}$ …(1)

Here,

K_p is the equilibrium constant in terms of partial pressures.

K is the equilibrium constant in terms of concentrations.

T is the temperature.

R is the universal gas constant.

$\mathrm{\Delta n}$is the numerical difference between the coefficients of gaseous products and the coefficients of gaseous reactants.

The value of$\mathrm{\Delta n}$ for the reaction $\mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{B}\left(\mathrm{g}\right)$can be calculated as follows:

$\begin{array}{c}\mathrm{\Delta n}=2 - \left(1+1\right)\\ = 0\end{array}$

Therefore, the K_p is equal to K because the role="math" localid="1664190196583" $\mathit{\Delta }\mathit{n}$value for this reaction is zero.

${\mathrm{K}}_{\mathrm{p}}=\mathrm{K}$

Hence, the value of K_p at 45^oC for the reaction is 0.579.

Find the value of x.

The equilibrium constant, K_p expression in terms of partial pressures is given below:

${\mathrm{K}}_{\mathrm{p}}=\frac{{{\mathrm{P}}_{\mathrm{B}}}^2}{\left({\mathrm{P}}_{\mathrm{C}}\right)\left({\mathrm{P}}_{\mathrm{D}}\right)}$ …(2)

Now construct the ICE table:

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{D}\left(\mathrm{g}\right) \\ \mathrm{Iinitial}\left(\mathrm{atm}\right)1.501.500 \\ \mathrm{Change}\left(\mathrm{atm}\right)- \mathrm{x}-\mathrm{x}+2\mathrm{x} \\ -------------------------- \\ \mathrm{Equilibrium}\left(\mathrm{atm}\right)1.50 -\mathrm{x}1.50 -\mathrm{x}2\mathrm{x} \end{gathered}$$

Substitute these equilibrium terms and the equilibrium constant into Equation (2) to find the value of x.

data-custom-editor="chemistry" $0.579=\frac{{\left(2\mathrm{x}\right)}^2}{\left(1.50 - \mathrm{x}\right)\left(1.50 - \mathrm{x}\right)}$

Solve for x.

$\mathrm{x}=0.413$

Find the equilibrium partial pressures of each species.

The equilibrium partial pressures of each speciesare as follows:

$$\begin{gathered} \begin{array}{c}{\mathrm{P}}_{\mathrm{C}}=1.50 - \mathrm{x}\\ =1.50 - 0.413\\ =1.087 \mathrm{atm}\end{array} \\ \begin{array}{c}{\mathrm{P}}_{\mathrm{D}}=1.50 - \mathrm{x}\\ =1.50 - 0.413\\ =1.087 \mathrm{atm}\end{array} \\ \begin{array}{c}{\mathrm{P}}_{\mathrm{B}}=2\mathrm{x}\\ =2\times 0.413\\ =0.826 \mathrm{atm}\end{array} \end{gathered}$$

Calculate the mole fraction of B once equilibrium reached.

The total partial pressure,$P_{total}$is the sum of the partial pressures of all the species.

${\mathrm{P}}_{\mathrm{total}}=1.087 \mathrm{atm}+1.087 \mathrm{atm}+0.826 \mathrm{atm}=3.00 \mathrm{atm}$

Use the below equation to calculate the mole fraction of B as follows:

$\begin{array}{c}{\mathrm{X}}_{\mathrm{B}}=\frac{{\mathrm{P}}_{\mathrm{B}}}{{\mathrm{P}}_{\mathrm{total}}}\\ =\frac{0.826 \mathrm{atm}}{3.00 \mathrm{atm}}\\ =0.275\end{array}$

Therefore, the mole fraction of B once equilibrium reached is 0.275.