The equilibrium constant K for the reaction H₂(g) + - Br₂(g) = 2 HBr (g) is 2.180 x 106 Part: 0 / 2 Part 1 of 2 at 730.0 °C. Starting with 2.20 moles of HBr in a 19.4 L reaction vessel, calculate the concentrations of H₂, Br₂, and HBr at equilibrium. Be sure each of your answer entries has the correct number of significant digits. [H₂] = 2.92 × 10 Br₂] = 2.92 × 10 M M x10 X

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HBr=.....M

The equilibrium constant K for the reaction
С
H₂(g) + Br₂(g) = 2 HBr (g)
is 2.180 × 106 at 730.0 °C. Starting with 2.20 moles of HBr in a 19.4 L reaction vessel, calculate the concentrations of H₂, Br₂, and HBr at equilibrium.
Part: 0 / 2
Part 1 of 2
Be sure each of your answer entries has the correct number of significant digits.
[H₂]
[Br₂]
= 2.92 × 10
= 2.92 × 10
M
M
x10
X
Ś
Transcribed Image Text:The equilibrium constant K for the reaction С H₂(g) + Br₂(g) = 2 HBr (g) is 2.180 × 106 at 730.0 °C. Starting with 2.20 moles of HBr in a 19.4 L reaction vessel, calculate the concentrations of H₂, Br₂, and HBr at equilibrium. Part: 0 / 2 Part 1 of 2 Be sure each of your answer entries has the correct number of significant digits. [H₂] [Br₂] = 2.92 × 10 = 2.92 × 10 M M x10 X Ś
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