2.6.2Equilibrium

Law of mass action and Kc, Kp

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WHAT is the Law of Mass Action?

For an elementary reaction aA+bBcC+dDaA + bB \rightarrow cC + dD: rateforward=kf[A]a[B]b\text{rate}_{forward} = k_f [A]^a [B]^b

WHY the powers? Because a reaction event needs aa molecules of A and bb of B to meet simultaneously. The chance of that meeting scales like [A]a[B]b[A]^a[B]^b — same logic as multiplying probabilities of independent events.


HOW we derive KcK_c (kinetic route)

Consider the reversible reaction: aA+bBcC+dDaA + bB \rightleftharpoons cC + dD

Step 1 — Forward rate (why: A,B collide to form products): rf=kf[A]a[B]br_f = k_f [A]^a [B]^b

Step 2 — Backward rate (why: C,D collide to reform reactants): rb=kb[C]c[D]dr_b = k_b [C]^c [D]^d

Step 3 — At equilibrium the rates are equal (why: no net change means forward = backward): kf[A]a[B]b=kb[C]c[D]dk_f [A]^a [B]^b = k_b [C]^c [D]^d

Step 4 — Rearrange (why: gather the two rate constants, both constant at fixed TT): kfkb=[C]c[D]d[A]a[B]b=Kc\frac{k_f}{k_b} = \frac{[C]^c [D]^d}{[A]^a [B]^b} = K_c


HOW we get KpK_p (for gases)

For gases it is easier to measure partial pressures. Define with partial pressures: Kp=pCcpDdpAapBbK_p = \frac{p_C^{\,c}\, p_D^{\,d}}{p_A^{\,a}\, p_B^{\,b}}

Relate KpK_p to KcK_c using the ideal gas law pV=nRTpV = nRT:

Step 1 — For each gas, partial pressure (why: pressure ∝ concentration for ideal gas): pi=niVRT=[i]RTwhere [i]=niVp_i = \frac{n_i}{V}RT = [i]\,RT \quad\text{where } [i]=\tfrac{n_i}{V}

Step 2 — Substitute pi=[i]RTp_i = [i]RT into KpK_p: Kp=([C]RT)c([D]RT)d([A]RT)a([B]RT)bK_p = \frac{([C]RT)^c\,([D]RT)^d}{([A]RT)^a\,([B]RT)^b}

Step 3 — Separate the RTRT powers (why: collect the exponents of RTRT): Kp=[C]c[D]d[A]a[B]bKc(RT)(c+d)(a+b)K_p = \underbrace{\frac{[C]^c[D]^d}{[A]^a[B]^b}}_{K_c}\cdot (RT)^{(c+d)-(a+b)}

Step 4 — Let Δng=(c+d)(a+b)\Delta n_g = (c+d)-(a+b) = moles of gaseous products − moles of gaseous reactants:

Figure — Law of mass action and Kc, Kp

Important rules (WHY they hold)

Manipulating equilibria:

  • Reverse a reaction → K=1/KK' = 1/K (numerator and denominator swap).
  • Multiply coefficients by nnK=KnK' = K^n (each term raised to nn).
  • Add two reactions → K=K1×K2K = K_1 \times K_2 (rates chain multiply).

Worked examples


Common mistakes


Flashcards

Law of mass action states rate ∝ ?
product of active masses of reactants, each raised to its stoichiometric coefficient
KcK_c for aA+bBcC+dDaA+bB\rightleftharpoons cC+dD
Kc=[C]c[D]d[A]a[B]bK_c=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}
Kinetic origin of KcK_c
at equilibrium kf[A]a[B]b=kb[C]c[D]dk_f[A]^a[B]^b=k_b[C]^c[D]^d, so Kc=kf/kbK_c=k_f/k_b
Relation between KpK_p and KcK_c
Kp=Kc(RT)ΔngK_p=K_c(RT)^{\Delta n_g}
Definition of Δng\Delta n_g
gaseous moles of products − gaseous moles of reactants
When does Kp=KcK_p=K_c?
when Δng=0\Delta n_g=0
Why omit pure solids/liquids from K
their activity (concentration) is constant, absorbed into K
Effect of catalyst on K
none — speeds both directions equally
Only factor that changes K
temperature
Reverse a reaction, new K = ?
1/K1/K
Multiply all coefficients by n, new K = ?
KnK^n
KpK_p for CaCO3(s)CaO(s)+CO2(g)CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)
Kp=pCO2K_p=p_{CO_2}

Recall Feynman: explain to a 12-year-old

Imagine a busy playground swap: kids trade marbles for stickers and back again. After a while the rate of trading marbles-to-stickers equals the rate of stickers-to-marbles. Nothing looks like it's changing, but trades still happen. If you count "how many stickers per marble" in the crowd, you always get the same number as long as the weather (temperature) stays the same. That fixed number is KK. Cool marbles (solids) sitting on the ground don't count — only the ones actively floating in the air (gases/dissolved) matter.

Connections

Concept Map

gives

gives

at equilibrium equals

rearrange

equals ratio

constant at fixed T

p = i RT

for gases use pressures

separate RT powers

linked by

delta n = 0

Law of Mass Action

Forward rate kf A^a B^b

Backward rate kb C^c D^d

Kc = products over reactants

kf over kb

Depends only on Temperature

Ideal gas law pV=nRT

Substitute pressures

Kp = partial pressures ratio

Kp = Kc RT^delta n

Kp equals Kc

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab koi reversible reaction hota hai (matlab dono taraf ja sakta hai), to shuru me forward reaction fast hoti hai aur backward slow. Dheere-dheere forward slow aur backward fast hoti jaati hai, aur ek point aata hai jahan dono ki rate barabar ho jaati hai — isko equilibrium kehte hain. Yahan par products aur reactants ka ek fixed ratio ban jaata hai, aur wahi ratio equilibrium constant KcK_c hai. Formula: products upar, reactants neeche, aur har ek ki power uska stoichiometric coefficient.

Iski derivation simple hai — law of mass action bolta hai rate ∝ concentration raised to coefficient. Forward rate =kf[A]a[B]b=k_f[A]^a[B]^b, backward =kb[C]c[D]d=k_b[C]^c[D]^d. Equilibrium par dono equal, to kf/kb=Kck_f/k_b = K_c. Kyunki dono kk constant hain fixed temperature par, KcK_c bhi constant. Sirf temperature hi KK ko badalta hai — pressure, concentration ya catalyst nahi.

Gases ke liye hum partial pressure use karte hain, to KpK_p banta hai. Ideal gas se pi=[i]RTp_i=[i]RT daalne par milta hai Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}, jahan Δng\Delta n_g = gaseous product moles minus gaseous reactant moles. Agar Δng=0\Delta n_g=0 to Kp=KcK_p=K_c. Yaad rakho — pure solid aur liquid ko KK me kabhi mat likho, unki concentration constant hoti hai. Yeh concept exam me har year aata hai, numericals easy hote hain agar formula aur Δng\Delta n_g ka sign sahi samajh lo.

Go deeper — visual, from zero

Test yourself — Equilibrium

Connections