For the formation of ammonia, the equilibrium constant is calculated to be 5.2×10^-5 at 25°C. After analysis, it is determined that [N2]=2.00M and [H2]=0.80M. How many grams of ammonia are in the 10L reaction vessel at equilibrium? Use the following equilibrium equation.
N2(g)+3H2(g)<--->2NH3(g)
Chemistry help?
2018-05-30 11:13 am
回答 (1)
2018-05-30 2:05 pm
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) …… Kc = 5.2 × 10⁻⁵
Kc = [NH₃]² / ([N₂] [H₂]³)
5.2 × 10⁻⁵ = [NH₃]² / (2.00 × 0.80³)
[NH₃] = √{(5.2 × 10⁻⁵) × 2.00 × 0.80³} M = 0.0073 M
No. of moles of NH₃ = (0.0073 mol/L) × (10 L) = 0.073 mol
Molar mass of NH₃ = (14.0 + 1.0×3) g/mol = 17.0 g/mol
Mass of NH₃ = (0.073 mol) × (17.0 g/mol) = 1.24 g ≈ 1.2 g (to 2 sig. fig.)
Kc = [NH₃]² / ([N₂] [H₂]³)
5.2 × 10⁻⁵ = [NH₃]² / (2.00 × 0.80³)
[NH₃] = √{(5.2 × 10⁻⁵) × 2.00 × 0.80³} M = 0.0073 M
No. of moles of NH₃ = (0.0073 mol/L) × (10 L) = 0.073 mol
Molar mass of NH₃ = (14.0 + 1.0×3) g/mol = 17.0 g/mol
Mass of NH₃ = (0.073 mol) × (17.0 g/mol) = 1.24 g ≈ 1.2 g (to 2 sig. fig.)
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