5.20 mol of solid A was placed in a sealed 1.00-L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.40 M, where it remained constant.
A(s)----B(g)+C(g)
Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain?
Chemistry problem. Help meeee please!!!!?
2016-06-21 12:03 pm
回答 (1)
2016-06-21 12:12 pm
✔ 最佳答案
A(s) ⇌ B(g) + C(g)Original equilibrium :
[B] = [C] = 1.40 mol/L
Kc = [B][C] = 1.40² = 1.96
New equilibrium when the container volume is doubled, i.e. V = 2.00 L :
As temperature is unchanged, Kc = 1.96 (remains unchanged)
Kc = [B][C] = 1.96
Then [B] = [C] = √1.96 mol/L = 1.40 mol/L
Number of moles of A reacted = (1.40 mol/L) × (2.00 L) = 2.80 mol
No. of moles of A remains = (5.20 - 2.80) mol = 2.40 mol
收錄日期: 2021-04-18 15:09:34
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