Respuesta :
Answer:
K = 3.37
Explanation:
2 NH₃(g) → N₂(g)  + 3H₂(g)
Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).
       2 NH₃(g)   →   N₂(g)  + 3H₂(g)
Initally    4moles       -       -
React     2moles      2m  +  3m
Eq       2 moles      2m     3m
We had produced 2 moles of nitrogen and 3 mol of Hâ‚‚ (ratio is 2:3)
The expression for K is:  ( [H₂]³ . [N₂] ) / [NH₃]²
We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)
K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²
K = 27/8 / 1 → 3.37
Answer:
The value of K for this reaction is 1.69
Explanation:
Step 1: Data given
Moles of NH3 = 4.0 moles
Volume of the container = 2.0 L
At the equilibrium 2.0 moles NH3 remains
Step 2: The balanced equation
2 NH3(g) → N2(g) + 3H2(g)
Step 3: Initial number of moles
NH3: 4.0 moles
N2: 0 moles
H2: 0 moles
Step 4: Number of moles at the equilibrium
NH3: 2.0 moles
This means there reacts 2.0 moles of NH3
For 2 moles of NH3 we have 1 mol of N2 and 3 moles of H2
There will be produced 1 mol of N2 and 3 moles of H2
Step 5: Calculate molarity
Molarity = moles / volume
Molarity of NH3 = 2.0 moles / 2.0 L = 1 M
Molarity of N2 = 1.0 mol / 2.0 L = 0.5 M
Molarity of H2 = 3.0 mol / 2.0 L = 1.5 M
Kc = ([H2]³[N2]) / [NH3]²
Kc = (1.5³ * 0.5) / (1²)
Kc = 1.69
The value of K for this reaction is 1.69
