Question

$2NOBr(g) \rightleftharpoons 2NO(g) + B{r_2}(g)$. If nitrosyl bromide (NOBr) is 40% dissociated at a certain temperature and a total pressure of 0.30 atm. ${K_p}$ for the reaction $2NO(g) + B{r_2}(g) \rightleftharpoons 2NOBr(g)$ is:
A. 45
B. 25
C. 0.025
D. 0.25

Answer 1

If a reactant has initial concentration let's say, P , then the pressure is created on the products according to their stoichiometry. The pressure of reactants decreases from their initial value at the equilibrium point. The sum of the pressure of the reactant and the product at equilibrium is called equilibrium pressure.

Complete step by step answer: ${K_p}$ is the equilibrium constant calculated from the partial pressures of the reaction equation . It is used to express the relationship between the product pressures and reactant pressures. It is a unit less number, although it relates to pressure.
Consider the equation (given in equation) that is the reciprocal of the equation:
$2NOBr(g) \rightleftharpoons 2NO(g) + B{r_2}(g)$
At time t=0 a 0 0
Time = t $a - a\alpha$ $\dfrac{{2a\alpha }}{2}$ $\dfrac{{a\alpha }}{2}$
$\alpha$ is 0.4 as 40% dissociation is occurring in the question.
As we know that the value of equilibrium constant in the terms of pressure can be written as:
$\dfrac{{{K_p} = {{({P_{NO}})}^2}({P_{B{r_2}}})}}{{({P_{NOBr{)^2}}}}}$
When we write the partial pressures of NO, $B{r_2}$ and NOBr, it can be written as follows:
Partial pressure of NO = $\dfrac{{a\alpha }}{{a(1 + \dfrac{\alpha }{2})}} \times P = \dfrac{{0.4}}{{1.2}} = 0.3$
= 0.1
Partial pressure of $B{r_2}$ = $\dfrac{{a\alpha }}{{a(1 + \dfrac{\alpha }{2})}} \times P = \dfrac{{0.4}}{{1.2}} = 0.05$
Partial pressure of NOBr =
$\dfrac{{a(1 - \alpha )}}{{a(1 + \dfrac{\alpha }{2})}} \times P = 0.15$
${K_p} = \dfrac{{{{(0.1)}^2} \times 0.05}}{{{{(0.15)}^2}}} = \dfrac{1}{{45}}$
This is ${K_p}$ value of the equation
$2NOBr(g) \rightleftharpoons 2NO(g) + B{r_2}(g)$
But as per the question, we need to find the value of $2NO(g) + B{r_2}(g) \rightleftharpoons 2NOBr(g)$
So the value of ${K_p}$ for this equation is
${({K_p})_{new}} = \dfrac{1}{{\dfrac{1}{{45}}}} = 45$
So, the correct answer is “Option A”.

Note:
KP and KC can be made related to each other using an equation.
${K_p} \to$ Equilibrium constant in terms of pressure.
${K_c} \to$ Equilibrium constant in terms of concentration.
${K_p} = {K_c}{(RT)^{\Delta ng}}$
$\Delta ng$= number of moles of gaseous product – number of moles of gaseous reactants