A Question about Chemical Equilibrium

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The discussion focuses on the equilibrium reaction N2O4 <====> 2NO2 and explores how to express changes in moles and pressure over time. It confirms that at equilibrium, the number of moles of N2O4 can be represented as 'a - x' and for NO2 as '2x', where 'x' is the amount of N2O4 that dissociates. Similarly, for pressure, the initial pressure 'p' can be expressed as 'p - y' for N2O4 and '2y' for NO2 at equilibrium, with 'y' representing the decrease in pressure of the reactant. The discussion also raises the question of whether x equals y, affirming that the change in moles is directly related to the change in pressure. This method is deemed valid for all reversible reactions, regardless of stoichiometric coefficients.
vijayramakrishnan
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say we have a reaction:

N2O4 <====> 2NO2


for this reaction,if initial number of moles of N2O4 is 'a' can we write

t = 0 a -
at equilbrium a-x 2x

similarly can we do the same for pressure i.e.
if initial pressure is p,then can we say

t=0 p -
at equilbrium p-y 2y

if yes why? and is x=y,i.e is the number of moles x dissociated equal to decrease in pressure of reactant?
and is this method valid for all reversible reaction irrespective of the stoichiometric coefficient of reactants and products

please help.
 
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Physics Forums - Equilibrium N2O4(g) <=> 2NO2(g)

------------------- N2O4(g) ---- <=> --- 2NO2(g)
Ci: ------------------ a ------------------------ 0 ---
∆C:---------------- -x ---------------------- +2x ---
Ceq: ------------- a - x -------------------- 2x ---

Kc = [NO2]2/[N2O4] = (2x)2/(a – x) = 4x2/(a – x) => Kc(a – x) = 4x2 => Kc·a - Kc·x = 4x2

4x2 + Kc·x + (-Kc·a) = 0

x = [-b ± (b2 – 4ac)1/2]/2a

a = 4, b = Kc , c = -Kc·a

substitute into quadratic; solve for x …

[N2O4] = a – x

[NO22] = 2x
 
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