Question

Given, $2NOBr \rightleftharpoons 2NO + B{r_2}$ If nitrosyl bromide (NOBr) is 34% dissociated at 25$^ \circ C$ & total pressure of 0.25atm. Calculate ${K_p}$ ​for the dissociation at this temperature.

Answer 1

We should start the question by taking initial pressure of NOBr at P. Then we will relate the dissociation of reactants and product with pressure. And by using formula for ${K_p}$ we can easily calculate it.

Complete answer:
We should know that while calculating ${K_p}$, the partial pressures of gases are used. The partial pressures of pure solids and liquids are not included. To use this equation, it is beneficial to have an understanding of partial pressures and mole fractions.
Partial Pressures: Dalton's law states that all of the partial pressures add up to the total pressure, as shown in the equation:
${P_{total}} = {P_A} + {P_B} + .......$
Individual gases maintain their respective pressures when combined. However, their individual pressures add up to the total pressure in the system
We should first take a look at the reaction:
$2NOBr \rightleftharpoons 2NO + B{r_2}$
This reaction is taking place at 25$^ \circ C$ and total pressure of 0.25 atm.
If the initial pressure of NOBr is p, then at equilibrium, the percent dissociation of NOBr is 34%. Then the partial pressures of :
P(NOBr) = 0.66p
P(NO) = 0.34p
P($B{r_2}$) = $\dfrac{{0.34}}{2}p$ = 0.17p
The total pressure of the solution is given in the solution to be = 0.25 atm
Total pressure = P(NOBr) + P(NO) + P($B{r_2}$)
0.25 = 0.66p + 0.34p + 0.17p
0.25 = 1.17p
So, p = $\dfrac{{0.25}}{{1.17}}$ = 0.214 atm
-We will now calculate the equilibrium partial pressures of all:
P(NOBr) = 0.66p = 0.66 × 0.214 = 0.141 atm
P(NO) = 0.34p = 0.34 × 0.214 = 0.073 atm
P($B{r_2}$) = 0.17p = 0.17 × 0.214 = 0.036 atm
Now we will use the original balance equation to calculate the value of ${K_p}$:
${K_p}$ = $\dfrac{{{{\left( {{P_{NO}}} \right)}^2}\left( {{P_{B{r^2}}}} \right)}}{{{{\left( {{P_{NOBr}}} \right)}^2}}}$
= $\dfrac{{{{\left( {0.073} \right)}^2}\left( {0.036} \right)}}{{{{\left( {0.141} \right)}^2}}}$
= 0.00964 atm
Hence the value of ${K_p}$ for this dissociation will be 0.00964 atm.

Note:
We should know that ${K_p}$ is the equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is a unit less number, although it relates the pressures.