Question

Nitric oxide and bromine at initial partial pressures of 98.4 torr and 41.3 torr, respectively, were allowed to react at 300. K. At equilibrium the total pressure was 110.5 torr. The reaction is

$\mathbf{2}\mathbf{NO}\left(\mathbf{g}\right)\mathbf{+}{\mathbf{Br}}_2\left(\mathbf{g}\right)\to \mathbf{2}\mathbf{NOBr}\left(\mathbf{g}\right)$

a. Calculate the value of${\mathrm{K}}_{\mathrm{p}}$.

b. What would be the partial pressures of all species if NOand,${\mathbf{Br}}_2$both at an initial partial pressure of 0.30 atm, were allowed to come to equilibrium at this temperature?

Short answer

a.$K_p=133.76atm$

b. Thus, the partial pressure of,$\text{NOBr}$, $\text{NO}$and ${\text{Br}}_{\text{2}}$are 0.248 atm, 0.052 atm, and 0.176 atm.

Step-by-step solution

Determine the value of Kp

The equilibrium reaction is as shown.

$\text{2NO}\left(\text{g}\right){\text{+Br}}_{\text{2}}\left(\text{g}\right)\to \text{2NOBr}\left(\text{g}\right)$

The equilibrium expression for the reaction is

$K_p=\frac{{\left[{\text{P}}_{NOBr}\right]}^2}{{\left[P_{NO}\right]}^2\left[P_{B{r}_2}\right]}$

The ICE table for the reaction is

$\begin{array}{ccccccc}& & 2NO\left(g\right)& +& B{r}_2\left(g\right)& \to & 2NOBr\left(g\right)\\ Initial\left(torr\right)& & 98.4& & 41.3& & 0\\ Change\left(torr\right)& & -2x& & -x& & +2x\\ AtEquilibrium\left(torr\right)& & 98.4-2x& & 41.3-x& & 2x\end{array}$

On substituting the given values, we get

$\begin{array}{c}{K}_p=\frac{{\left[{\text{P}}_{NOBr}\right]}^2}{{\left[P_{NO}\right]}^2\left[P_{B{r}_2}\right]}\\ =\frac{4x^2}{{(98.4-2x)}^2(41.3-x)}\end{array}$

The total pressure at equilibrium is 110.5 torr. Because of this, we can assume the following as

$\begin{array}{c}{P}_{total}=\sum P_{partial}\\ 110.5=2x+(98.4-2x)+(41.3-x)\end{array}$

By calculating the value of x, and on substituting it in the$K_p$expression, we get

$\begin{array}{c}110.5=2x+98.4-2x+41.3-x\\ 110.5=139.7-x\\ x=139.7-110.5\\ =29.2\end{array}$

On substituting it in the$K_p$expression, we get

$\begin{array}{c}{K}_p=\frac{{\left[{\text{P}}_{NOBr}\right]}^2}{{\left[P_{NO}\right]}^2\left[P_{B{r}_2}\right]}\\ =\frac{4x^2}{{(98.4-2x)}^2(41.3-x)}\\ =\frac{4\times {\left(29.2\right)}^2}{{(98.4-2\left(29.2\right))}^2(41.3-29.2)}\\ =\frac{3410.4}{{\left(40\right)}^2\left(12.1\right)}\end{array}$

On simplifying the given equation, we get

$\begin{array}{c}{K}_p=\frac{3410.4}{1600\times 12.1}\\ =\frac{3410.4}{19360}\\ =0.176\end{array}$

By converting torr to atm as follows:

$\begin{array}{c}{K}_p=\frac{0.176}{torr}\times \frac{760torr}{1atm}\\ K_p=133.76atm\end{array}$

Determine the value of partial pressure

b.

The ICE table for the reaction is as shown.

$\begin{array}{ccccccc}& & 2NO\left(g\right)& +& B{r}_2\left(g\right)& \to & 2NOBr\left(g\right)\\ Initial\left(atm\right)& & 0.30& & 0.30& & 0\\ Change\left(atm\right)& & -2x& & -x& & +2x\\ AtEquilibrium\left(atm\right)& & 0.30-2x& & 0.30-x& & 2x\end{array}$

Upon substituting the given values, we get

$\begin{array}{c}{K}_p=\frac{{\left[{\text{P}}_{NOBr}\right]}^2}{{\left[P_{NO}\right]}^2\left[P_{B{r}_2}\right]}\\ 133.76=\frac{4x^2}{{(98.4-2x)}^2(41.3-x)}\end{array}$

Upon simplifying the given equation, we get

X=0.124

Finally, to determine the partial pressure of all species, we can substitute the x value in for each pressure at equilibrium as:

$\begin{array}{c}{P}_{NOB{r}_2}=2\times x\\ =2\times 0.124\\ =0.248atm\end{array}$

$\begin{array}{c}{P}_{NO}=(0.30-2x)\\ =(0.30-2\times 0.124)\\ =(30-0.248)\\ =0.052atm\end{array}$

$\begin{array}{c}{P}_{B{r}_2}=(0.30-x)\\ =(.30-0.124)\\ =0.176atm\end{array}$

Thus, the partial pressure of,$\text{NOBr}$, $\text{NO}$and ${\text{Br}}_{\text{2}}$are 0.248 atm, 0.052 atm, and 0.176 atm respectively.