Question

At 2200^oC, K = 0.050 for the reaction

${\mathbf{\text{N}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g)+O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\stackrel{}{\mathbf{\rightleftharpoons }}\mathbf{\text{2NO(g)}}$

What is the partial pressure of NO at equilibrium assuming the N₂ and O₂ had initial pressures of 0.80 atm and 0.20 atm, respectively?

Short answer

The equilibrium partial pressure of NO at the given temperature is 0.078 atm.

Step-by-step solution

The relation between  Kp and Kc

The relationship between K_p and K can be written as follows:

${\text{K}}_{\text{p}}\text{=K}{\left(\text{RT}\right)}^{\text{Δn}}$ …(1)

Here,

K_p is the equilibrium constant in terms of partial pressures.

K is the equilibrium constant in terms of concentrations.

R is the universal gas constant.

T is the temperature.

$\text{Δn}$is the numerical difference between the coefficients of gaseous products and the coefficients of gaseous reactants.

The value of $\text{Δn}$for the given reaction can be determined as shown below:

$\begin{array}{c}\text{Δn=2\hspace{0.17em}–\hspace{0.17em}}\left(\text{1+1}\right)\\ \text{=0}\end{array}$

When$\text{Δn=0}$, then Equation (1) becomes,

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

Since the term,${\left(\text{RT}\right)}^{\text{0}}$ is equal to 1.

Hence, the value of K_p for the reaction is 0.050.

The expression for equilibrium constant

The equation for the equilibrium constant, K_p of the reaction in terms of partial pressures is written below:

${\text{K}}_{\text{P}}\text{=}\frac{{{\text{P}}_{\text{NO}}}^{\text{2}}}{\left({\text{P}}_{{\text{N}}_{\text{2}}}\right)\left({\text{P}}_{{\text{O}}_{\text{2}}}\right)}$ …(2)

The given initial partial pressures of${\text{N}}_{\text{2}}$ and${\text{O}}_{\text{2}}$ are 0.80 atm and 0.20 atm, respectively.

ICE table

Assume that x is the change in partial pressure.

Setup the ICE table as:

$\begin{array}{cccccc}& {\text{N}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{O}}_{\text{2}}\left(\text{g}\right)& \stackrel{}{\rightleftharpoons }& \text{2NO}\left(\text{g}\right)\\ \text{Iinitial}\left(\text{atm}\right)& \text{0.80}& & \text{0.20}& & \text{0}\\ \text{Change}\left(\text{atm}\right)& \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{+2x}\\ \text{Equilibrium}\left(\text{atm}\right)& \text{0.80\hspace{0.17em}–\hspace{0.17em}x}& & \text{0.20\hspace{0.17em}–\hspace{0.17em}x}& & \text{2x}\end{array}$

Calculation

Insert the K_p value and the terms of the equilibrium partial pressures of the gases from ICE table into equation (2) to find the value of x.

$\text{0.050=}\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{0.80\hspace{0.17em}–\hspace{0.17em}x}\right)\left(\text{0.20\hspace{0.17em}–\hspace{0.17em}x}\right)}$

Solve for x.

$\text{x=0.039}$

The equilibrium partial pressure of NO gas

The equilibrium partial pressure of NOcan be calculated as shown below:

$\begin{array}{c}{\text{P}}_{\text{NO}}\text{=2x}\\ \text{=2×0.039}\\ \text{=0.078\hspace{0.17em}atm}\end{array}$

Therefore, the equilibrium partial pressure of NO is 0.078 atm.