The equilibrium constant of a reaction is 300. If the volume of the reaction flask is tripled, the Equilibrium constant will be
(a) 100
(b) 900
(c) 600
(d) 300
Answered Feb 05, 2023
(d) The equilibrium constant will be 300.
(d) :- For those reactions, where Δn = 0, increase in volume at constant temperature does not affect the number of moles at equilibrium.
(2) Let the initial length and breadth be l and b, respectively. So, the initial perimeter = 2 (l + b) The final perimeter = 2 (2l + 2b) =...
P = 5 x 106 Pa = 5 x 106 / 105 bar n = 20 T = 300 K R = 0.083 L bar K-1 mol-1 Vreal = 800 mL Videal = nRT/P Putting values ,...
∆G =0 because the reaction is in equilibrium. ∆G0 =- 2.303RT log K = - 2.303X8.314JK-1 mol-1 x300K log 102 = -11488J mol-1 = -11.488kJmol-1 .
Kc=1/4..........
(i) The state of equilibrium is not disturbed on adding catalyst rather it is attained quickly. (ii) No effect on state of equilibrium
As the value of equilibrium constant increases with increase in temperature so, the reaction is endothermic in nature.
V1=1litre V2=? T1= 270c=300k, T2=370C=310k At Constant pressure, V1T1 = V1/T1=V2/T2 1/300=V2/310 V2 = 1.0333 litre Since capacity of flask is one litre Volume of air escaped out = 1.0333-1= 0.0333 litre = 33.3ml
(c) The equilibrium constant (K) for the reaction is 0.05.
Three, because rate = k [A]3
Install the Bissoy app to consult with a doctor.
Log in to ask questions, provide answers, or leave comments.