Question

For the reaction a(g) 2 b(g), a reaction vessel initially contains only a at a pressure of pa = 1.32 atm. at equilibrium, pa = 0.25 atm. calculate the value of kp. (assume no changes in volume or temperature.)

Answer 1

When there is no change in volume and temperature:
So we can assume that:
P1/n1 = P2/n2
and p1/p2 = n1/n2
So we can assume that the coefficients in the balanced equation as pressure ratios = mole ratios
so from the reaction:
A(g) ↔ 2 B (g)
∴ 1 mole A → 2 mole B
when we have PA(initial) = 1.32 atm & PB (initial) = 0 atm & PA(equ) = 0.25 atm 
∴PB (intial) = (PA(intial))-PA(equ) * 2 mol
                  = (1.32 - 0.25) *2 = 2.14 atm

SO we can get Kp from this formula:
Kp = p(B)^2 / [P(A)
     = 2.14^2 / 0.25
     = 18.32

Answer 2

A gaseous reaction at equilibrium is defined not only by its molarity but also by its partial pressure. Therefore the coefficient Kp for the given reaction is 18.32.

Given here,

The reaction,

$\bold {A (g)\rightleftharpoons 2B(g)}$

1 mole of reactant  A forms 2 moles of product B.

The initial pressure of PA(i) = 1.32 atm

Pressure at equilibrium PA(equi) =  0.25 atm

At constant pressure and temperature

$\bold {\dfrac {P1}{n1} = \dfrac {P2}{n2}}$

or

$\bold {\dfrac {P1}{P2} = \dfrac {n1}{n2}}$

The initial pressure of the product

$\bold {PB (i) = PA(i)-PA(equi) \times 2 mol}$

Put the values in the equation

$\bold {PB (l) = (1.32 - 0.25) \times 2 }\\\\\bold {PB (i) = 2.14 atm}}$

Kp of the reaction can be calculated by the formula,

$\bold {Kp = \dfrac {P(B)^2} {P(A)}}$

put the values

$\bold {Kp = \dfrac {(2.14)^2} { 0.25}}\\\\\bold {Kp = 18.32}$

Therefore the coefficient Kp for the given reaction is 18.32.

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