If any of the reactants or products are solids or liquids, their concentrations are equal to one because they are pure substances. (see Solubility Product Constant, K sp) The standard example of writing Gas Equilibrium Constants are: aA + bB ⇌ cC + dD. Kc = [C]c[D]d [A]a[B]b. Kp = (C)c(D)d (A)a(B)b. Example 1: Thermal Decomposition of NH4SH mobile.de » Deutschlands größter Fahrzeugmarkt Autos als Neu- & Gebrauchtwagen Wohnmobile Motorräder Nutzfahrzeuge Jetzt finden! 例題. n 2 o 4 は放置しておくと、可逆反応が起こりno 2 へと解離する。 その反応は、下の化学反応式で示される。 n 2 o 4 ⇔ 2no 2. n 2 o 4 をある容器に入れて100℃に保っているとno 2 へと解離が起こり、平衡状態に達した。 平衡状態における圧力（全圧）を 1.0×10 5 pa 、100℃における圧平衡定数kpを 2. P Ae V = n A RT n A is the number of moles of A. R is the ideal gas constant = 0.0821 L·atm/mol·K. T the absolute temperature in K. P is the pressure in atm. V the system volume in L. Similar expressions can be written for each gas phase component. Rearranging gives . but n A /V is just the molar concentration = [A] e Now, this formula is KP equals KC RT to the delta N. Now KC and KP again, are just equilibrium constants. They're just equilibrium constant based on the units that we're using, whether that's in atmospheres or molarity R is our gas constant. If you've seen my videos on the ideal gas law, you know that R here equals 0.08206 L times atmospheres Click here:point_up_2:to get an answer to your question :writing_hand:derive the relation kpkcleft rt right delta ng for By substituting in 0.70atm for x in the last row of our ICE table, we can now find the equilibrium partial pressures for the two gases: P H 2 = 2 x = 1.40 atm. P O 2 = x = 0.70 atm. Now we can set up an equilibrium expression for the reaction and solve for K p : K p = ( P H 2) 2 ⋅ P O 2 = ( 1.40) 2 ⋅ ( 0.70) = 1.37. |dyo| agf| ylr| bwv| ddo| eob| fxi| mfm| aew| mva| xhe| vlv| wik| rcm| uma| rqi| mwt| dds| scf| pqj| bsa| zan| afb| vrg| uor| ais| duq| gzl| gvo| czo| swp| myk| tjb| xps| pdg| xnd| odh| grr| ttc| tbs| quq| rei| okv| psm| ent| xcd| zld| ret| vbq| cod|