Answer:
[H₂] = 6.74×10⁻³ M
[I₂] = 4.65×10⁻³ M
[HI] = 0.0413 M
Explanation:
The equilibrium is:
H₂(g) + I₂(g) ⇌ 2 HI(g)
Expression for Kc = [HI]² / [H₂] . [I₂] 54.3
We analyse the equation:
H₂(g) + I₂(g) ⇌ 2 HI(g)
Initially 0.00623 0.00414 0.0424
React x x 2x
x amount has reacted, therefore by stoichiometry 2x has been added to the HI, that we have in the beginning.
Eq 0.00623-x 0.00414-x 0.0424+2x
We make the expression for Kc:
Kc = (0.0424+2x)² / (0.00623-x) . (0.00414-x) = 54.3
This is quadractic funcion:
54.3 = (1.79×10⁻³ + 0.1696x + 4x²) / (2.58×10⁻⁵- 0.01037x + x²)
54.3 (2.58×10⁻⁵- 0.01037x + x²) = 1.79×10⁻³ + 0.1696x + 4x²
1.40×10⁻³- 0.563x + 54.3x² = 1.79×10⁻³ + 0.1696x + 4x²
-3.89×10⁻⁴ - 0.7326x +50.3x² = 0 → a = 50.3 ; b= - 0.7326 ; c = -3.89×10⁻⁴
Quadratic formula = (-b +- √(b² + 4ac))/ (2a)
x₁ = 0.015
x₂ = -5.13×10⁻⁴ . We choose x₂ as x₁ give us negative concentrations
[H₂] = 0.00623 - (-5.13×10⁻⁴) = 6.74×10⁻³ M
[I₂] = 0.00414 - (-5.13×10⁻⁴) = 4.65×10⁻ M
[HI] = 0.0424+2(-5.13×10⁻⁴) = 0.0413 M
The concentration will be "0.00663 M, 0.00454 M, and 0.0416 M".
According to the question,
Value of Kc = 54.3
Temperature = 430°C
The equation will be:
→ H₂ (g) + I₂ (g) [tex]\rightleftharpoons[/tex] 2HI (g)
Here,
Initial concentration,
[H2] =0.00623 M
[I2] = 0.00414 M
[HI] = 0.0424 M
Value of Kc = 54.3
We know the relation,
→ [tex]K_c[/tex] = [tex]\frac{[HI]^2}{[H_2][I_2]}[/tex]
By substituting the values,
54.3 = [tex]\frac{(0.0424+2x)^2}{(0.00623-x)(0.00414-x)}[/tex]
54.3x² + 0.0014 - 0.563x = 0.0017 + 4x² + 0.1696x
50.3x² - 0.732x - 0.0003 = 0
x = -0.0004
Now,
[H₂] equilibrium = 0.00623 + 0.0004
= 0.00663 M
[I₂] equilibrium = 0.00414 + 0.0004
= 0.00454 M
[HI] equilibrium = 0.0424 - 2(0.0004)
= 0.0416 M
Thus the response above is right.
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