Question #32cda
1 Answer
Explanation:
We're asked to calculate the
Let's first write the equilibrium constant expression for this reaction:
We can create a makeshift I.C.E. chart via bullet points as follows:
Initial Concentrations:

#"NO"_2# :#(0.240color(white)(l)"mol")/(2.00color(white)(l)"L") = 0.120M# 
#"NO"# :#0# 
#"O"_2# :#0#
because there are no
From the coefficients of the reaction equation, we expect the changes to be
Change in concentration:

#"NO"_2# :#2x# 
#"NO"# :#+2x# 
#"O"_2# :#+x#
Knowing that the equilibrium concentration of
We can calculate the actual change in concentration of
Which means
So the equilibrium concentrations are
Equilibrium Concentrations:

#"NO"_2# :#0.0895M# 
#"NO"# :#2(color(red)(0.01525M)) = 0.0305M# 
#"O"_2# :#color(red)(0.01525M#
We now plug these into the
Which agrees with our calculations that equilibrium lies to the left (more reactants present).