Chemical Equilibrium

NH₄OH $\rightleftharpoons$ NH$_{4}^+$ + OH^$-$

K_b = $\frac{[NH_{4}^+][OH^-]}{[NH_4OH]}$

1.8 $\times$ 10^$-$5 = $\frac{[NH_{4}^+][0.01]}{[0.02]}$

$\therefore$ [NH$_{4}^+$] = 3.6 $\times$ 10^$-$5 M

van't Hoff equation giving variation of equilibrium constant, K with temperature T is

log_e K = $-$ $\frac{\Delta H^\circ}{RT}+\frac{\Delta S^\circ}{R}$

Given, log_e K = $-$ $\frac{2000}{T}+4.0$

$\therefore$ $\Delta$S$^\circ$/R = 4 $\Rightarrow$ $\Delta$S$^\circ$ = 4R

Image

${K_C} = {{[{C^2}][D]} \over {[A][{B^2}]}} = {{{{(0.2)}^2}(0.1)} \over {(1){{(2)}^2}}} = 0.001$

H₂SO₄(aq) $\to$ H⁺(aq) + HSO$_4^ - $(aq); K_a1 = 1 $\times$ 10³

H₂SO$_4^ - $(aq) $\to$ H⁺(aq) + SO$_4^ - $(aq); K_a2 = 1.2 $\times$ 10^$-$2

Here, K_a2 < K_a1 because the negatively charged HSO$_4^ - $ ion has much less tendency to donate a proton to H₂O as compared to neutral H₂SO₄.