For the following reaction: H2 (g) + Br2 (g)→ 2HBr (g) platinum is used as catalyst.
How does the platinum lower the activation energy and therefore promote H2 and Br2 to react with eachother?

For the following reaction: H2 (g) + Br2 (g)→ 2HBr (g) platinum is used as catalyst. How does the platinum lower the act...
Consider the reaction 2HBr(g) H2(g) Br2(1) The standard free energy change for this reaction is 107.0 kJ. The free energy change when 2.50 moles of HBr(g) react at standard condition is kJ. What is the maximum amount of useful work that the reaction of 2.50 moles of HBr(g) is capable of producing in the surroundings under standard conditions? If no work can be done, enter none. kJ
Consider the reaction 2HBr(g) H2(g) Br2(1) The standard free energy change for this...
Consider the reaction: 2HBr(g)H2(g) + Br2(l) Using standard thermodynamic data at 298K, calculate the free energy change when 2.33 moles of HBr(g) react at standard conditions. G°rxn = _____kJ
H2C=CH2(g) + H2(g) C2H6(g) Adding a small amount of platinum powder greatly increases the rate of the reaction above. How might the platinum be affecting the rate of the reaction? (Select all that apply.) It makes the electron density in a bond lower so that it is easier to break. It orients the reactants in the proper direction. It increases the energy of activation for the reaction. It increases the concentration of the reactants. It acts as a catalyst in...
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
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] =
Consider the following reaction:
H2(g)+Br2(g)→2HBr(g)
The graph below shows the concentration of Br2 as a function of
time.(Figure 1)
Make a rough sketch of a curve
representing the concentration of HBr as a function of time. Assume
that the initial concentration of HBr is zero.
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
The equilibrium constant for the following reaction: H2(g) + Br2(g) ↔ 2HBr (g) is 1.20 x 103 at a certain temperature. Find the equilibrium pressure of HBr if 9.20 atm of HBr is introduced into a sealed container at this temperature.
The equilibrium constant for the following reaction: H2(g) + Br2(g) ↔ 2HBr (g) is 1.90 x 103 at a certain temperature. Find the equilibrium pressure of HBr if 14.50 atm of HBr is introduced into a sealed container at this temperature.