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a guest Mar 20th, 2018 34 Never
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  1. $$
  2. \begin{array}{ccc}
  3. &\ce{2NO2(g) &<=>& N2O4(g)}\\\hline
  4. -&a&&b&\\
  5. t=0&a+x&&b\\
  6. t=t_0&a+x-2y&&b+y\\
  7. \end{array}
  8. $$
  9.  
  10. At the first equilibrium, $K_\mathrm{p}=\frac{b}{a^2}$. At $t=0$, we add more $\ce{NO2}$, and let it reach equilibrium at $t=t_0$. All of $a,b,x,y$ are gas pressure in Pascals. Since the temperature is constant, the value of $K_\mathrm{p}$ remains the same at $t=t_0$.
  11.  
  12. So, we can say $$K_\mathrm{p}=\frac{b}{a^2}=\frac{b+y}{(a+x-2y)^2}$$
  13.  
  14. Rearranging we get an equation quadratic in both $x$ and $y$:
  15.  
  16. $$4by^2-y(4xb+4ab+a^2)+2abx+bx^2=0$$
  17.  
  18. Solving it for $y$ we get the two solutions:
  19.  
  20. $$y_1 = \frac{-a \sqrt{a^2 + 8 a b + 8 b (2 b + x)} + a^2 + 4 a b + 4 b x}{8 b}$$
  21.  
  22. and $$y_2 = \frac{a \sqrt{a^2 + 8 a b + 8 b (2 b + x)} + a^2 + 4 a b + 4 b x}{8 b}$$
  23.  
  24. Now, to prove that $a+x-2y>a$ I'm stuck :(
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