Respuesta :
Answer:
The answer to the question is;
The equilibrium constant for the reaction is 0.278
Reversibility.
Explanation:
Initial concentration = 0.500 M N₂ and 0.800 M H₂
N₂ (g) + 3·H₂ (g) ⇔ 2·NH₃ (g)
One mole of nitrogen combines with three moles of hydrogen form 2 moles of ammonia
That is 1 mole of ammonia requires 3/2 moles of H₂ and 1/2 moles of N₂
0.150 M of ammonia requires 3/2×0.150 moles of H₂ and 1/2×0.150 moles of N₂
That is 0.150 M of ammonia requires 0.225 moles of H₂ and 0.075 moles of N₂
Therefore at equilibrium we have
Number of moles of Nitrogen = 0.500 M - 0.075 M = 0.425 M
Number of moles of Hydrogen = 0.800 M - 0.225 M = 0.575 M
Number of moles of Ammonia = 0.150 M
K[tex]_c[/tex] = [tex]\frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.15)^2}{(0.425)*(0.575)^3}[/tex] = 0.278
The kind of reaction is a reversible one as the equilibrium constant is greater than 0.01 which as general guide, all components in a reaction with an equilibrium constant between the ranges of 0.01 and 100 will be present when equilibrium is reached and the chemical reaction will be reversible.
Answer:
The equilibrium constant for this reactions is 0.278
Explanation:
Step 1: Data given
Concentration of N2 = 0.500 M
Concentration of H2 = 0.800 M
At equilibrium the concentration of ammonia is 0.150 M
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: The initial concentrations
[N2] = 0.500 M
[H2] = 0.800 M
[NH3] = 0M
Step 4: The concentration at equilibrium
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
[N2] = (0.500 - X)M
[H2] = (0.800 - 3X)M
[NH3] = 2X M = 0.150 M
X = 0.075 M
[N2] = (0.500 - 0.075) = 0.425 M
[H2] = (0.800 - 3X)M = 0.575 M
[NH3] = 2X M = 0.150 M
Step 5: Calculate Kc
Kc = [NH3]² / [N2][H2]³
Kc = [0.150]² / [0.425][0.575]³
Kc = 0.278
The equilibrium constant for this reactions is 0.278
