The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K.
2HI(g) H2(g) + I2(g)
An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.319 M HI, 4.28×10-2 M H2 and 4.28×10-2 M I2.
What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.50×10-2 mol of H2(g) is added to the flask?
[HI] = M
[H2] = M
[I2] = M
Concentration changes at equilibrium?
2016-05-10 12:29 pm
回答 (1)
2016-05-10 1:56 pm
2HI(g) ⇌ H₂(g) + I₂(g) ...... K = 1.80 × 10⁻²
The equilibrium is disturbed when 3.50 × 10⁻² mol of H₂(g) is added :
[HI]ₒ = 0.319 mol/L
[H₂]ₒ = (4.28 × 10⁻² mol/L) + (3.50 × 10⁻² mol/L) = 7.78 × 10⁻² mol/L
[I₂]ₒ = 4.28 × 10⁻² mol/L
At new equilibrium :
Let y mol be the amount of H₂ consumed.
[HI] = (0.319 + y) mol/L
[H₂] = {(7.78 × 10⁻²) - y} mol/L
[I₂] = {(4.28 × 10⁻²) - y} mol/L
K = [H₂] [I₂] / [HI]²
1.80 × 10⁻² = {(7.78 × 10⁻²) - y} {(4.28 × 10⁻²) - y} / (0.319 + y)²
(1.80 × 10⁻²)y² + (1.1484 × 10⁻²) y + (1.831698 × 10⁻³) = y² - 0.1206y + (3.32984 × 10⁻³)
0.982y² - 0.132084y + (1.498142 × 10⁻³) = 0
y = {0.132084 ± √[0.132084² - 4 × 0.982 ×(1.498142 × 10⁻³)]} / (2 × 0.982)
y = 1.25 × 10⁻² or y = 0.122 (rejected for [H₂] < 0 and [I₂] < 0)
At new equilibrium :
[HI] = [0.319 + (1.25 × 10⁻²)] mol/L = 0.332 mol/L
[H₂] = {(7.78 × 10⁻²) - (1.25 × 10⁻²)} mol/L = 6.53 × 10⁻² mol/L
[I₂] = {(4.28 × 10⁻²) - (1.25 × 10⁻²)} mol/L = 3.03 × 10⁻² mol/L
The equilibrium is disturbed when 3.50 × 10⁻² mol of H₂(g) is added :
[HI]ₒ = 0.319 mol/L
[H₂]ₒ = (4.28 × 10⁻² mol/L) + (3.50 × 10⁻² mol/L) = 7.78 × 10⁻² mol/L
[I₂]ₒ = 4.28 × 10⁻² mol/L
At new equilibrium :
Let y mol be the amount of H₂ consumed.
[HI] = (0.319 + y) mol/L
[H₂] = {(7.78 × 10⁻²) - y} mol/L
[I₂] = {(4.28 × 10⁻²) - y} mol/L
K = [H₂] [I₂] / [HI]²
1.80 × 10⁻² = {(7.78 × 10⁻²) - y} {(4.28 × 10⁻²) - y} / (0.319 + y)²
(1.80 × 10⁻²)y² + (1.1484 × 10⁻²) y + (1.831698 × 10⁻³) = y² - 0.1206y + (3.32984 × 10⁻³)
0.982y² - 0.132084y + (1.498142 × 10⁻³) = 0
y = {0.132084 ± √[0.132084² - 4 × 0.982 ×(1.498142 × 10⁻³)]} / (2 × 0.982)
y = 1.25 × 10⁻² or y = 0.122 (rejected for [H₂] < 0 and [I₂] < 0)
At new equilibrium :
[HI] = [0.319 + (1.25 × 10⁻²)] mol/L = 0.332 mol/L
[H₂] = {(7.78 × 10⁻²) - (1.25 × 10⁻²)} mol/L = 6.53 × 10⁻² mol/L
[I₂] = {(4.28 × 10⁻²) - (1.25 × 10⁻²)} mol/L = 3.03 × 10⁻² mol/L
收錄日期: 2021-04-18 14:48:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160510042912AAf2dTe