### Related Formula
K_p = fracp_mathrmAB cdot p_mathrmB_2^1/2p_mathrmAB_2$$K_p = \frac{p_{\mathrm{AB}} \cdot p_{\mathrm{B}_2}^{1/2}}{p_{\mathrm{AB}_2}}$$
### Core Logic
Consider the equilibrium reaction setup:
| State | mathrmAB_2(g)$\mathrm{AB}_{2(g)}$ | rightleftharpoons$\rightleftharpoons$ | mathrmAB_(g)$\mathrm{AB}_{(g)}$ | +$+$ | frac12mathrmB_2(g)$\frac{1}{2}\mathrm{B}_{2(g)}$ |
|---|
| Initial moles: | 1$1$ | | 0$0$ | | 0$0$ |
| Equilibrium moles: | 1 - x$1 - x$ | | x$x$ | | fracx2$\frac{x}{2}$ |
Total equilibrium moles:
n_texttotal = 1 - x + x + fracx2 = 1 + fracx2$$n_{\text{total}} = 1 - x + x + \frac{x}{2} = 1 + \frac{x}{2}$$
Since
x ll 1$x \ll 1$, total moles
n_texttotal approx 1$n_{\text{total}} \approx 1$ and
(1 - x) approx 1$(1 - x) \approx 1$.
Partial pressures:
p_mathrmAB_2 approx p$$p_{\mathrm{AB}_2} \approx p$$
p_mathrmAB approx x p$$p_{\mathrm{AB}} \approx x p$$
p_mathrmB_2 approx fracx2 p$$p_{\mathrm{B}_2} \approx \frac{x}{2} p$$
Substituting into
K_p$K_p$:
K_p = frac(x p) cdot left(dfracx p2right)^1/2p = x cdot left(fracx p2right)^1/2 = fracx^3/2 p^1/2sqrt2$$K_p = \frac{(x p) \cdot \left(\dfrac{x p}{2}\right)^{1/2}}{p} = x \cdot \left(\frac{x p}{2}\right)^{1/2} = \frac{x^{3/2} p^{1/2}}{\sqrt{2}}$$
Squaring both sides and solving for
x$x$:
K_p^2 = fracx^3 p2 implies x^3 = frac2K_p^2p$$K_p^2 = \frac{x^3 p}{2} \implies x^3 = \frac{2K_p^2}{p}$$
x = sqrt[3]frac2K_p^2p$$x = \sqrt[3]{\frac{2K_p^2}{p}}$$
### Pattern Recognition
For
Delta n_g = 0.5$\Delta n_g = 0.5$ involving degree of dissociation
x ll 1$x \ll 1$, tracking total pressure approximations ensures an immediate analytical solution without full polynomial expansion.
### Chapter Mix
Class 11 Chemistry: Chemical Equilibrium
