Question

A sample of gaseous Nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at 25⁰C according to the following equation:

$\mathbf{2}\mathbf{NOBr}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{Br}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

The initial density of the system was recorded as 4.495 g/L. After equilibrium was reached, the density was noted to be 4.086 g/L.

a. Determine the value of the equilibrium constant K for the reaction.

b. If Ar(g) is added to the system at equilibrium at a constant temperature, what will happen to the equilibrium position? What happens to the value of K? Explain each answer.

Short answer

a. The value of the equilibrium constant K for the reaction is 2.33 x 10^-3

b. When the argon gas is added it will increase the volume of the container and Q < K, and the reaction shifts right to get back to equilibrium

Step-by-step solution

First, we have to calculate the Initial no. of moles of NOBr from the given value of density:

The given reaction is:$2\mathrm{NOBr}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Br}}_2\left(\mathrm{g}\right)$

Initial density = 4.495g/l, at equilibrium density = 4.086g/l

Let’s assume that the initial volume is 1 lit:

Since: $\mathbf{density}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{volume}}\mathbf{=}\mathbf{;}\mathbf{mass}\mathbf{=}\mathbf{volume}\mathbf{\times }\mathbf{density}$

The molecular mass of NOBr is $\left(109.92)\right)$

Initial moles of NOBr$=4.495\mathrm{g}/\mathrm{Lit}\times \frac{1\mathrm{mol}}{109.92\mathrm{gm}}\times 1\mathrm{Lit}=0.04089\mathrm{MOL}$

Now find out the volume at equilibrium:

Since: $\mathbf{density}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{volume}}$

role="math" localid="1657983605893" $\frac{\mathrm{initial}\mathrm{density}}{\mathrm{quilibrium}\mathrm{density}}=\frac{\frac{\mathrm{mass}}{\mathrm{initial}\mathrm{volume}}}{\frac{\mathrm{mass}}{\mathrm{equilibrium}\mathrm{volume}}}$

role="math" localid="1657983628710" $\frac{4.496\mathrm{g}/\mathrm{lit}}{4.086\mathrm{g}/\mathrm{lit}}=1.100=\frac{\mathrm{equilibrium}\mathrm{volume}}{1\mathrm{Lit}}$

Therefore, volume at equilibrium = 1.1 Lit

From the reaction: $2\mathrm{NOBr}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Br}}_2\left(\mathrm{g}\right)$

Initial 0.04090 mol → 0 + 0

Change −2x → +2x + x

Equilibrium(0.04090 – 2x)→+2x + x

role="math" localid="1657983720310" $\frac{\mathbf{initial}\mathbf{}\mathbf{volume}}{\mathbf{equilibrium}\mathbf{}\mathbf{volume}}\mathbf{=}\frac{\mathbf{no}\mathbf{.}\mathbf{}\mathbf{of}\mathbf{}\mathbf{initial}\mathbf{}\mathbf{moles}\mathbf{\left(}{\mathbf{n}}_{\mathbf{i}}\mathbf{\right)}}{\mathbf{no}\mathbf{.}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{}\mathbf{at}\mathbf{}\mathbf{equilibrium}\mathbf{\left(}{\mathbf{n}}_{\mathbf{e}}\mathbf{\right)}}$

$\frac{1\mathrm{Lit}}{1.1\mathrm{Lit}}=\frac{0.04089}{(0.040\mathrm{mol}-2\mathrm{x})+2\mathrm{x}+\mathrm{x}}\mathrm{Mol}$

$\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00409}\mathbf{mol}$

Now solve for the equilibrium concentration:

$\left[\mathrm{NOBr}\right]=\frac{0.03272\mathrm{mol}}{1.1\mathrm{Lit}}=0.02975\mathrm{M}$

$\left[\mathrm{NO}\right]=\frac{0.0818\mathrm{mol}}{1.1\mathrm{Lit}}=0.00744\mathrm{M}$

$\left[\mathrm{Br}\right]=\frac{0.0409\mathrm{mol}}{1.1\mathrm{Lit}}=0.00372\mathrm{M}$

Now equilibrium constant is:

$\mathbf{K}\mathbf{=}\frac{{\mathbf{\left[}\mathbf{NO}\mathbf{\right]}}^{\mathbf{2}}\mathbf{\left[}{\mathbf{Br}}_{\mathbf{2}}\mathbf{\right]}}{{\mathbf{\left[}\mathbf{NOBr}\mathbf{\right]}}^{\mathbf{2}}}$

$\mathrm{K}=\frac{{(0.00744)}^2(0.00372)}{{(0.02975)}^2}$

Equilibrium constant(K) = 2.33 x 10^-4

b.

As we know, the volume increases at constant temperature and pressure when any gas is added to the container. Therefore, when the argon gas is added, it will increase the volume of the container.

This is because the container is a constant-pressure system, and if the number of moles increases at constant T and P, the volume must increase.

An increase in volume will dilute the concentrations of all gaseous reactants and gaseous products.

Because there are more moles of product gases versus reactant gases

(3 mol versus 2 mol), the dilution will decrease the numerator of K more than the denominator will decrease.

This causes Q < K, and the reaction shifts right to get back to equilibrium.

Because the temperature was unchanged, the value of K will not change. K is a constant as long as the temperature is constant.