Ammonia is produced via #N_2 (g) + 3H_2 rightleftharpoons 2NH_3 (g)#. During the production process, the production engineer determines the reaction quotient to be Q = #3.56xx10^-4#. lf K= #6.02xx10^-2#, what can be said about the reaction?
1 Answer
The reaction proceeds in the forward direction.
Explanation:
The idea here is that you need to compare the value of the reaction quotient,
You know that at a certain temperature, the equilibrium
#"N"_ (2(g)) + 3"H"_ (2(g)) rightleftharpoons 2"NH"_ (3(g))#
has an equilibrium constant equal to
#K_c = 6.02 * 10^(-2)#
The equilibrium constant is calculated using equilibrium concentrations. In this case, you can say that
#K_c = (["NH"_3]^2)/(["N"_2] * ["H"_2]^3)#
Now, the reaction quotient is calculated using the the concentrations of the chemical species that take part in the reaction measured at a specific point in the progress of the reaction.
This implies that you can use the reaction quotient to predict if the reaction is at equilibrium. You can thus distinguish three possible cases
#Q_c < K_c# In this case, the equilibrium will shift in such a way as to increase the concentration of the product and decrease the concentrations of the reactants.
This implies that the forward reaction will be favored.
#Q_c = K_c# In this case, the reaction is at equilibrium, i.e. the forward and the reverse reaction proceed at the same rate.
#Q_c > K_c# When this happens, the equilibrium will shift in such a way as to decrease the concentration of the product and increase the concentrations of the reactants.
As a result, the reverse reaction will be favored.
In your case,
#3.56 * 10^(-4) < 6.02 * 10^(-2)#
You can thus say that forward reaction will be favored
#"N"_ (2(g)) + 3"H"_ (2(g)) -> 3"NH"_ (3(g))#
As a result, the reaction will consume nitrogen gas and hydrogen gas, your reactants, and produce ammonia, your product.
The reaction will proceed in the forward direction until equilibrium is established. At that point, the forward and the reverse reactions will proceed at the same rate, meaning that neither the forward, nor the reverse reaction is favored.
