For the reaction H2+Br2 <--> 2HBr . Kc= 2.18×10^8. Calculate the values of H2 and Br at equilibrium if we start with only HBr present.. HBr= 3.20M
For the reaction H2+Br2 <--> 2HBr . Kc= 2.18×10^8. Calculate the values of H2 and Br...
4) The equilibrium constant kc for the reaction H2(g) + Br2(g) = 2HBr(g is 2.18 x 106 at 730°C. Starting with HBr only with (HBr]° = 0.267 M, calculate the concentrations of H2, Brz, and HBr at equilibrium.
The Kc of the reaction H2 (g)+Br2 (g)=2HBr (g) is 2.18×10+6. If the initial concentration of HBr in 12.0L vessel is 3.20 moles, calculate the concentration of H2, Br2 and HBr at equilibrium. Use ICE table. Show calculations and all work & may have to use quadratic equation.
10. The equilibrium constant Kc for the reaction H2(g) + Br2(g) ⇆ 2HBr(g) is 2.180 × 106 at 730°C. Starting with 2.20 moles of HBr in a 13.7−L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium. [H2] = [Br2] = [HBr] =
The equilibrium constant Kc for the reaction H2(g) + Br2(g) ⇆ 2HBr(g) is 2.180 × 106 at 730°C. Starting with 1.20 moles of HBr in a 21.3−L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium.
The equilibrium constant K. for the reaction H2(g) + Brz(g) = 2HBr(g) is 2.18 x 106 at 730°C. Starting with 3.20 moles of HBr in a 12.0-L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium. (10 points) (Reference: Chang 14.43
The value of the equilibrium constant for the reaction 2HBr(g)<....>H2(g)+Br2(g) is Kc=1.26*10^-12 at 500k. A. what would be the value of the equilibrium constant K'c for the related reaction written in the following fashion? 1/2 H2(g)+1/2 Br2(g)<....>HBr(g). B. what will be the corresponding value for Kp, the pressure form of the equilibrium constant? (R=0.08206)
Be sure to answer all parts. The equilibrium constant Kc for the reaction H2(g) + Br2(g) ⇆ 2HBr(g) is 2.180 × 106 at 730°C. Starting with 2.20 moles of HBr in a 18.1−L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium. [H2] =___ M Br2] = ___M [HBr] = ____M
The molar equilibrium constant, Kc, is 7.7x10-11at 25oC for the reaction: 2HBr(g) <--> H2(g)+ Br2(g) What is the gas-phase equilibrium constant, Kp, for the reaction: Br2(g)+ H2(g) <--> 2HBr(g) a. 1.8x10-9 b. 7.7x10-11 c. 0.0 d. 1.3x1010 e. 3.8x1011
For the reaction: H2+Br2 ⇌ 2HBr, Kc = 7.5x10^2 at a certain temperature. 1.00 mole HBr is placed in 5.0 L flask at a certain temperature. What is the concentration of HBr at equilibrium? a) 0.19 b) 0.81 c) 0.01 d) 0.94 e) 0.03 What I got so far for the molarity, H2 + Br2 ⇌ 2HBr I 0 0 .2 C +x +x -2x (due to 2 moles) E x x .2-x ----------------------------------------------- for kc 7,5x10^2 = 750, put...
5. For the overall exothermic reaction: H2(g) + Br2(g) → 2HBr(g) the following mechanism was determined: fast Equilibrium Step 1: Brz(8) 2Br(g) Step 2: H2(g) + Br(g)_52HBr(g) + H(g) Step 3: H(g) + Br(g) k3 HBr(g) slow fast Use a plot of AH versus Reaction Pathway to illustrate the three step reaction profile.