For the reaction
4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g) ,
K = 10^80 at a certain temperature. Initially, all reactants and products have concentrations equal to 12 M. What is the approximate concentration of ammonia at equilibrium?
a. 3 M
b. 12 M
c. 18 M
d. 6 M
e. 0 M
To find the approximate concentration of ammonia at equilibrium, we need to consider the equilibrium constant (K) and the stoichiometry of the reaction.
The equilibrium constant (K) is a measure of the extent to which a reaction proceeds towards the products at equilibrium. In this case, the value of K is given as 10^80.
By looking at the balanced chemical equation, we can determine the stoichiometry of the reaction. According to the balanced equation:
4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g)
For every 4 moles of NH3, 2 moles of N2 are formed. Therefore, the stoichiometric coefficient ratio of NH3 to N2 is 4:2, or simplified as 2:1.
Since the initial concentration of all reactants and products is given as 12 M, we can assume that the change in concentration is negligible compared to the initial concentration. Therefore, the concentration of ammonia (NH3) at equilibrium can be approximated as half of the initial concentration, which is 6 M.
Thus, the approximate concentration of ammonia at equilibrium is 6 M, which corresponds to answer choice (d).
To summarize, we used the stoichiometry of the reaction and the equilibrium constant to calculate the approximate concentration of ammonia at equilibrium.