For the reaction H2(g) + Br2(g) 2HBr(g), Kc = 81.4 at 385 °C. If [H2] = [Br2] =
ID: 947916 • Letter: F
Question
For the reaction H2(g) + Br2(g) 2HBr(g), Kc = 81.4 at 385 °C. If [H2] = [Br2] = 0.24 M and [HBr] = 2.4 M at 385 °C. which one of the following is correct?
[H2] and [HBr] decrease as the system moves toward equilibrium.
The system is at equilibrium.
[H2] and [Br2] increase as the system approaches equilibrium.
[HBr] increases as the system approaches equilibrium.
[HBr] and [Br2] increase as the system approaches equilibrium.
A.[H2] and [HBr] decrease as the system moves toward equilibrium.
B.The system is at equilibrium.
C.[H2] and [Br2] increase as the system approaches equilibrium.
D.[HBr] increases as the system approaches equilibrium.
E.[HBr] and [Br2] increase as the system approaches equilibrium.
Explanation / Answer
Reaction is,
H2 + Br2 -----> 2 HBr
Kc = 81.4
[H2] = [Br2] = 0.24 M
[HBr] =2.4 M
We know that,
Qc = [HBr]^2 / [H2] [Br2]
=> Qc = (2.4)^2 / 0.24 x 0.24 = 100
We see that,
Qc > Kc.
Therfore, the system is not at equilibrium.
To decrease the value of Qc till it reaches Kc, reaction will move in the backward direction.
The correct answer would be,
[H2] and [Br2] increase as the system approaches equilibrium
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