Please explain the process! Determine if the equilibrium lies to the right or left for the...
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7. The equilibrium constant at a particuiar temperature for the reaction: H2(g) + Br2(g) -> 2HBr(g) is 1.5 x 105 The value of Kc for HBr(g) -1 /2 H2(g) + 1 /2 Br2(g) is PCla(g) + Cl2(g) For the equilibrium: PCIs(g)-- Kc = 4.0 at 228°C. If 0.35 g of pure PC15 is added to a 2.0 L flask what will be the equilibrium concentration of Cl2(g). (you'll need to carry extra sig figs to...
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] =
Please explain how you got the answer for this, for some reason I can not grasp how to choose from those answer choices... BUT I do know how to get the Qc... Thanks so much for your help and please explain as simply as possible For the reaction H2(g) + Br2(g) 2HBr(g), Kc = 81.4 at 385C. If [H2] = [Br2] = [HBr] = 2.4 10-4 M at 385C, which one of the following is correct? A. [H2] and [HBr]...
35) Given the equilibrium reaction at constant pressure: 2HBr(g) + 72.7 kJ = H2(g) + Br2(g) When the temperature is increased, the equilibrium will shift to the A) left, and the concentration of HBr(g) will decrease B) right, and the concentration of HBr(g) will decrease C) right, and the concentration of HBr(g) will increase D) left, and the concentration of HBr(g) will increase
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.
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
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26. Consider the reaction: 2HBr(g) = H2(g) + Br2(g) Keq = 49.0 If 1.20 M H2, 1.20 M Br2 and 1.00 M HBr are placed into a container at 44° C, which of the following is true as equilibrium is approached? A. [Br2] decreases significantly. B. [HBr] decreases significantly. C. [H2] decreases significantly. D. [H2) remains the same. 27. Which Keq is most likely to favour the formation of reactants?...
The equilibrium constant for: F2(g) + 2HBr(g) ↔ 2HF + Br2 , is 1.0 x 10-5. The equilibrium lies: A. to the left (mostly reactants) B. to the right (mostly products) C. in the middle (50% reactants, 50% products)
For the following reaction, Kp = 2.8 ✕ 104 at 1630 K. H2(g) + Br2(g) equilibrium reaction arrow 2 HBr(g) What is the value of Kp for the following reactions at 1630 K? (a) HBr(g) equilibrium reaction arrow 1/2 H2(g) + 1/2 Br2(g) (b) 2HBr(g) equilibrium reaction arrow H2(g) + Br2(g) (c) 1/2H2(g) + 1/2 Br2(g) equilibrium reaction arrow HBr(g)
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Show your work when there are calculations, write units where appropriate, and use the correct significant figures. Do the work on a separate sheet of paper and once you have done it correctly, transcribe youir work clearly onto this sheet. 1. The value of Ke for the equilibrium: is 794 at 25°C a) At this temperature, what is the value of Ke for the following equilibrium? H2(g) + 12(g)% 2H1(g) HI(g)与½ H2(g) + ½ 12(g) b)...