At 2000°C the equilibrium constant for the reaction is Kc = 2.4 ✕ 103.
2 NO(g) equilibrium reaction arrow N2(g) + O2(g)
If the initial concentration of NO is 0.160 M, what are the equilibrium concentrations of NO, N2, and O2?
I do not even know how to approach this question.
You do it all with an ICE chart.
...........2NO ==> N2 + O2
I........0.160.....0.....0
C..........-2x.....x.....x
E.......0.160-2x...x.....x
Substitute the E line into the Keq expression and solve for x and 0.150-x.
Post your work if you get stuck.
No worries! I'll break it down for you step by step.
To solve this question, we'll use the concept of equilibrium and the equilibrium constant (Kc). The equilibrium constant gives us a measure of the extent to which a reaction has reached equilibrium. It is calculated using the concentrations of the reactants and products at equilibrium.
First, let's write down the balanced equation for the reaction:
2 NO(g) ⇌ N2(g) + O2(g)
Given: Kc = 2.4 ✕ 10^3
Initial concentration of NO = 0.160 M
To determine the equilibrium concentrations of NO, N2, and O2, you'll need to use an ICE table, which stands for Initial, Change, and Equilibrium.
Step 1: Fill in the Initial row of the ICE table:
Species | Initial Concentration
NO | 0.160 M
N2 | 0 M (since it is not provided initially)
O2 | 0 M (since it is not provided initially)
Step 2: Fill in the Change row of the ICE table:
Species | Initial Concentration | Change in Concentration
NO | 0.160 M | -2x (because 2 moles of NO react to form 1 mole of N2)
N2 | 0 M | +x (since the reaction produces N2)
O2 | 0 M | +x (since the reaction produces O2)
Step 3: Fill in the Equilibrium row of the ICE table:
Species | Initial Concentration | Change in Concentration | Equilibrium Concentration
NO | 0.160 M | -2x | 0.160 - 2x
N2 | 0 M | +x | x
O2 | 0 M | +x | x
Step 4: Write the expression for the equilibrium constant (Kc) using the equilibrium concentrations of the species:
Kc = [N2][O2] / [NO]^2
Step 5: Substitute the equilibrium concentrations into the Kc expression:
Kc = (x)(x) / (0.160 - 2x)^2
Step 6: Solve for x:
Since the value of Kc is given as 2.4 ✕ 10^3, you can substitute this value and solve the equation:
2.4 ✕ 10^3 = (x)(x) / (0.160 - 2x)^2
Using algebraic techniques, you can rearrange the equation, solve for x, and obtain the equilibrium concentrations of NO, N2, and O2.
Note: The calculations involved in solving for x and the equilibrium concentrations may be quite involved, so I recommend using a numerical solver or a graphing calculator to obtain the values.
I hope this explanation helps you understand how to approach this type of question! Let me know if you have any further questions.