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Derive using steady state approximation for both H ans Br atoms. H2(g) + Br2(g) -> 2HBr(g)
steady state approximation rate law for this reaction: k1 Br2 <--> 2Br k-1 (this refers to reverse of Br2 to 2Br) k2 2Br + H2 ----> 2HBr
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.
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
Consider the reaction: 2HBR(g) >H2(g) + Br2() Using standard thermodynamic data at 298K, calculate the free energy change when 1.51 moles of HBr(g) react at standard conditions AG° kJ rxn AHof (kJ/mol) AG°F (kJ/mol) s° (J/mol K) Beryllium Вe(s) 0 9.5 -569.0 ВeO(s) -599.0 14.0 Be(ОН)2(s) -902.5 -815.0 51.9 AH°f (kJ/mol) AG°f (kJ/mol) s° (J/mol K) Bromine Br(g) 111.9 175.0 82.4 Br2() 152.2 0 0 Br2(g) 30.9 3.1 245.5 Br2(aq) -3.0 4.0 130.0 -121.0 -175.0 82.0 Br (aq) BrF3(g) -255.6...
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] =
6. Given the following data: Brig) 2 Brig) ΔΗ'-192.5 kJ/mol H2(g) + Br2(g)-) 2HBr(g) ΔΜ"--72.4 k/mol Calculate ΔΗ' for the reaction below H(g) + Br(g) - HBr(g)
For the reaction 2HBr(g) H2(g) +. Br2(1) H' = 72.6 kJ and Δ So=-114.5 J/K The equilibrium constant for this reaction at 295.0 K is Assume that Δ11° and Δ Sa are independent of temperature. Submit Answer more group attempt ramaining
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 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(g) + Br2(g) → 2HBr(g) Kp = 3.6 x 104 at 1494 K. What is the value of Kp for the following reaction at 1494 K? 42 H2(g) + / Br2(g) HBr(g) K". p Submit