The chemical system will shift to the left in order to produce more N₂ , H₂.
The missing data in the question is
The equilibrium constant (K) of the reaction below is K = 6.0 x 10⁻² with initial concentrations as follows: H₂ = 1.0 x 10⁻² M, N₂ = 4.0 M, and NH₃ = 1.0 x 10⁻⁴M.
<h3>What is Chemical Equilibrium?</h3>When the rate of forward reaction is equal to the rate of the backward reaction then the equation is said to be in Chemical Equilibrium.
N₂ + 3H₂ --> 2NH₃
The reaction quotient, Q, has the same algebraic expressions but uses the actual concentrations of reactants.
[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M
Thus, for the reaction:
N₂ + 3H₂ ⇄ 2NH₃
The equilibrium constant, K, of this reaction, is defined as:
Q = (5.6 x 10⁻³ )² / (1.0 x 10⁻²) ³(4.0 )
Q = 7.84
As Q > K,
The chemical system will shift to the left in order to produce more N₂ , H₂
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Answer:
The empirical formula is the simplest form;
Given:
Oxygen O at 94.1% and
H at 5.9%
Assume 100grams.
94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O
5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H
There is one mole of O for each mole of H so the empirical formula is
and written as OH.