2.6.3
Gas-phase reaction ke liye hum equilibrium constant ko concentrations (K c partial pressures (K p usi equilibrium ko describe karte hain, isliye dono ka aapas mein connection hona chahiye. In dono ke beech ka bridge ideal gas law hai, jo pressure aur concentration ko connect karta hai. Yeh link sirf ==Δ n
Ek general gas reaction a A + b B ⇌ c C + d D K c = [ A ] a [ B ] b [ C ] c [ D ] d , K p = p A a p B b p C c p D d
[ X ] p X X Δ n = ( c + d ) − ( a + b ) (gaseous products ke moles) − (gaseous reactants ke moles)
Δ n
Δ n Relationship
Matlab
Δ n = 0 K p = K c dono sides gas ke equal moles
Δ n > 0 K p > K c R T > 1 zyada gas moles produce hote hain
Δ n < 0 K p < K c gas moles consume hote hain
Worked example Example 1 — Ammonia synthesisN 2 ( g ) + 3 H 2 ( g ) ⇌ 2 N H 3 ( g ) K c = 0.5 K p
Step 1: Δ n = 2 − ( 1 + 3 ) = − 2 Kyun? Gas moles count karo: 2 product, 4 reactant.
Step 2: K p = K c ( R T ) Δ n = 0.5 × ( 0.0821 × 400 ) − 2 Kyun? Directly substitute karo; negative exponent note karo kyunki gas moles decrease ho rahe hain.
Step 3: R T = 32.84 K p = 0.5/ ( 32.84 ) 2 = 0.5/1078 = 4.6 × 1 0 − 4 Kyun chota? Δ n < 0 R T > 1 ( R T ) Δ n < 1 K p < K c
Worked example Example 2 — Decomposition (Δ n > 0 P C l 5 ( g ) ⇌ P C l 3 ( g ) + C l 2 ( g ) K c = 2.0 × 1 0 − 2
Step 1: Δ n = 2 − 1 = + 1 Step 2: K p = 2.0 × 1 0 − 2 × ( 0.0821 × 500 ) 1 Kyun? Gas ka ek extra mole banta hai, exponent + 1 Step 3: R T = 41.05 K p = 0.02 × 41.05 = 0.821 K p > K c
Worked example Example 3 —Δ n = 0 H 2 ( g ) + I 2 ( g ) ⇌ 2 H I ( g ) K c = 49
Δ n = 2 − 2 = 0 ⇒ ( R T ) 0 = 1 ⇒ K p = K c = 49 Kyun? Gas moles mein koi change nahi, to pressure aur concentration formulations ek hi hain. ✔
Δ n Galat soch: "C a C O 3 ( s ) ⇌ C a O ( s ) + C O 2 ( g ) Δ n = 2 − 1 = + 1 Kyun sahi lagta hai: Tum equation ki har species count karte ho.
Fix: Sirf gases count hoti hain. Yahan Δ n = 1 − 0 = + 1 C O 2 K
T Galat: Room temperature ke liye T = 27 Fix: Kelvin mein convert karo: T = 27 + 273 = 300 K
Δ n Galat soch: Δ n = Fix: Hamesha products − reactants . Yaad rakho: K Δ n
Common mistake Bhool jaana kiK p / K c R K R
What is the relationship between Kp and Kc? K p = K c ( R T ) Δ n
K p = K c ( R T ) Δ n Δ n Gaseous products ke moles minus gaseous reactants ke moles (products − reactants).
K p = K c ( R T ) Δ n Ideal gas law, p X = [ X ] R T
K p = K c Jab Δ n = 0 ( R T ) 0 = 1
N 2 + 3 H 2 ⇌ 2 N H 3 Δ n 2 − 4 = − 2
Δ n Sirf gaseous species; solids aur liquids exclude hote hain.
Temperature ki unit kya honi chahiye? Kelvin.
atm aur mol/L ke liye R ki value? 0.0821 L atm K − 1 mol − 1
Agar Δ n > 0 R T > 1 K p K c Bada (K p > K c
p X p X V = n X R T ⇒ p X = ( n X / V ) R T = [ X ] R T
Recall Feynman: ek 12-saal ke bache ko samjhaoSocho ek reaction ek box ke andar ho rahi hai jo gases se bhari hai. Tum describe kar sakte ho ki har gas "kitni bheedi" hai do tarike se: ya to har litre mein kitne molecules hain (concentration), ya wo walls par kitna zor lagate hain (pressure). Ideal gas law kehta hai dono same cheez hain, bas R T R T R T Kitne R T R T
"Products Push Past Reactants" → Δ n = "P for Pressure, P for Products first" taaki sign sahi rahe. Poora formula: Kp = Kc times RT to the change in gas moles.
Concentration = nX over V
Kp from partial pressures
Delta n = product gas moles - reactant gas moles
