Respuesta :
Answer:
a) 0.11
b)76.9
c) 8.8
d) 1.7*10^-4
Explanation:
Step 1: Data given
K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 2: Formula of K
aA(g) + bB(g) ⇌ cC(g) + dD(g)
K = [C]^c *[D]^d / [A]^a * [B]^b
K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]
Step 3:
a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
1/2N2 + 3/2H2(g) ⇌ NH3(g) =>K' = [tex]\sqrt{K}[/tex]
K' = [tex]\sqrt{1.3*10^-2}[/tex] = 0.11
b. 2NH3(g) ⇌ N2(g) + 3H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
2NH3(g) ⇌ N2(g) + 3H2(g) =>K' = 1/K
K' = 1/(1.3*10^-2) = 76.9
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
=>K' = [tex]\frac{1}{\sqrt{K} }[/tex]
K' = [tex]\frac{1}{\sqrt{1.3*10^-2} }[/tex]
K' = 8.8
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
2N2(g) + 6H2(g) ⇌ 4NH3(g)
K' = K²
K' = (1.3*10^-2)²
K' = 1.7 *10 ^-4
Values of equilibrium constant at given temperature for the following reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.
How we calculate equilibrium constant?
Equilibrium constant is define as the ration of the concentrations of product to the concentrations of reactant with respect to the exponent of their coefficients.
Given chemical reaction is:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Equilibrium constant for this reaction is:
K = [NH₃]² / [N₂][H₂]³
K = 1.3 × 10⁻² (given)
- Equilibrium constant K₁ for below reaction will be written as:
1/2N₂(g) + 3/2H₂(g) ⇌ NH₃(g)
K₁ = √K
Because concentration of all given species is 1/2 of the given reaction, so value of K₁ will be written as:
K₁ = √(1.3 × 10⁻²) = 0.11
- Equilibrium constant K₂ for below reaction will be written as:
2NH₃(g) ⇌ N₂(g) + 3H₂(g)
K₂ = 1/K
Because concentration of reactant and products are reciprocal from the concentration of original given reaction, so value of K₂ will be written as:
K₂ = 1/1.3 × 10⁻² = 76.9
- Equilibrium constant K₃ for below reaction will be written as:
NH₃(g) ⇌ 1/2N₂(g) + 3/2H₂(g)
K₃ = 1/√K
Because concentrations of given species is reciprocal as well as half of the given original reaction, so value of K₃ will be written as:
K₃ = 1/√(1.3 × 10⁻²) = 8.8
- Equilibrium constant K₄ for below reaction will be written as:
2N₂(g) + 6H₂(g) ⇌ 4NH₃(g)
K₄ = K²
Because concentrations of given species is double of the given original reaction, so value of K₄ will be written as:
K₄ = (1.3 × 10⁻²)² = 1.7 × 10⁻⁴
Hence, the value of K for given reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.
To know more about equilibrium constant, visit the below link:
https://brainly.com/question/12858312
