Respuesta :
Answer:
The equilibrium constant ≈120.07
Explanation:
Step 1: The balanced equation
N2(g)+3H2(g)↽−−⇀2NH3(g)
This means that for 1 mole of N2 consumed, there is needed 3 moles of H2 to produce 2 moles of NH3
Step 2: Calculate initial moles
Reactants:
⇒N2: 0.4011 mol es
⇒H2: 1.621 mol es
Products
⇒NH3: 0 mol
Step 3: Calculate number of moles that changed
N2: 0.4011 mol - 0.1001 mol = 0.3010 mol
H2: 3 * 0.3010 mol = 0.9030 mol
NH3: 2 * 0.3010 mol = 0.6020 mol
Step 4: Calculate moles at the equilibrium
N2: 0.1001 moles
⇒[N2] = 0.1001moles / 3.5L = 0.0286 M
H2: 1.621 moles -0.9030 moles = 0.718 moles
⇒[H2] = 0.718 moles/ 3.5L : 0.205 M
NH3: 0.6020 mol es
⇒[NH3] = 0.6020 / 3.5= 0.172M
Step 5: Calculate Kc
Kc = [NH3]² / ([N2] [H2]³)
Kc= 0.172²/(0.0286*0.205³)
Kc = 0.029584/(0.0286* 0.008615125)
Kc ≈120.07
The Kcis a high value, this shows the position of the equilibrium is to the right.
The equilibrium constant ≈120.07
The equilibrium constant, Kc is 120.07.
What is Kc?
Kc is the ratio of equilibrium product concentrations to equilibrium reactant concentrations.
The equation is
N₂(g) + 3H₂ = 2NH₃(g)
1 mole of nitrogen and 3 moles of hydrogen are needed to produce 2 moles of ammonia.
Step1: Calculating the initial moles
Reactants
N₂ = 0.4011 moles
H₂ = 1.621 moles
Products
NH₃ = 0 mol
Step2: Calculating the number of moles
N₂ = 0.4011 mol - 0.1001 mol = 0.3010 mol
H₂ = 3 × 0.3010 mol = 0.9030 mol
NH₃= 2× 0.3010 mol = 0.6020 mo
Step3: calculating the moles at the equilibrium
N₂ = 0.1001 moles
[N₂] = [tex]\dfrac{ 0.1001moles }{ 3.5L } = 0.0286\;M[/tex]
[H₂] = 1.621 moles - 0.9030 moles = 0.718 moles
[tex][H2] = \dfrac{0.718 moles}{3.5L} = 0.205 M[/tex]
NH₃ = 0.6020 moles
[tex][NH3] = \dfrac{0.6020 }{3.5L} = 0.172M[/tex]
Step4: Calculating the Kc
[tex]Kc = \dfrac{[NH3]² }{[N2] [H2]³}[/tex]
[tex]Kc= \dfrac{0.172²}{(0.0286 × 0.205³)}[/tex]
[tex]Kc = \dfrac{0.029584}{(0.0286 × 0.008615125)}[/tex]
Kc = 120.07
Thus, The equilibrium constant =120.07
Learn more about Kc, here:
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