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cheyenne481
18.06.2020 •
Chemistry
The following reaction was performed in a sealed vessel at 734 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.75M and [I2]=2.15M . The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature? Express your answer numerically.
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Ответ:
128
Explanation:
Step 1: Write the balanced equation
H₂(g) + I₂(g) ⇌ 2 HI(g)
Step 2: Make an ICE chart
H₂(g) + I₂(g) ⇌ 2 HI(g)
I 3.75 2.15 0
C -x -x +2x
E 3.75-x 2.15-x 2x
Step 3: Find the value of x
Since the concentration of I₂ at equilibrium is 0.0800 M,
2.15M-x = 0.0800 M
x = 2.07 M
Step 4: Calculate the concentrations at equilibrium
[H₂] = 3.75-x = 3.75-2.07 = 1.68 M
[I₂] = 0.0800 M
[HI] = 2x = 2(2.07) = 4.14 M
Step 5: Calculate the equilibrium constant Kc
Ответ:
HF + NaOH = NaF + H₂O
HF + Na⁺ + OH⁻ = Na⁺ + F⁻ + H₂O
HF + OH⁻ = F⁻ + H₂O