WebFor the reaction 3A (g) k → B(g) + C (g), k is 10−4 Lmol−1min−1. If [A] = 0.5 M then the value of −d[A] dt (in M s−1) is. Q. For the reaction 2A+B→C+D,−d[A] dt =k[A]2[B]. The … WebConsider the equilibrium reaction 3A+B↽−−⇀2C3A+B↽−−⇀2C Write the equation for the reverse reaction. reverse reaction: 2C↽−−⇀3A+B If the equilibrium concentrations are …

Evaluation of the Anticancer Activity and Mechanism Studies of

WebFor the reaction A+3B⇌2C+D, initial mole of A is twice that of B. If at equilibrium moles of B are equal, then percent of B reacted is: A 10% B 20% C 40% D 60% Medium Solution Verified by Toppr Correct option is D) Given reaction is A+3B⇌2C+D t=0 , a a/2 0 0 t=eqm , a−x a/2−3x 2x x At equilibrium, 2a−3x=2x (Given) 2a=5x⇒x= 10a WebJun 11, 2024 · Consider the reaction 3A + B + C ---> D + E where the rate law is defined as -D [A]/Dt=k [A]^2 [B] [C]? The initial concentration of B and C are both 1.00 M. The initial concentration of A is 1.00 ×x10−4M, and … signature bank grayscale

For the reaction 3 A + 4 B → 2 C + D, what is the …

WebFor the reaction, \[\ce{3A + 2B -> C + D}\], the differential rate law can be written as `underline(- 1/3 ("d"["A"])/"dt" = ("d"["C"])/"dt" = k ["A"]^n ["B"]^m)`. WebTranscribed Image Text: A reaction has the stoichiometry: 3A + B → C+ D. The following data were obtained for the initial rate of formation of C at various concentrations of A and B. Initial Concentration [A] 0.10 0.20 0.20 [B] 0.10 Hint 0.20 0.10 Hint Initial Rate of Formation of C (mol L-1 s-¹) Rate = k [A]⁰ [B]² None of these. WebThe equilibrium reaction is 3A(g)+B(g) ⇌ 2C (g). Let V be the volume of the flask. The equilibrium number of moles of A,B and C are 2,2 and 2 respectively. The equilibrium concentrations of A,B and C are V 2, V 2 … signature bank law firm