Question

The reaction in the interactive worked example is carried out at a different temperature at which kc = 0.055. this time, however, the reaction mixture starts with only the product, [no] = 0.0100 m, and no reactants. find the equilibrium concentrations of n2, o2, and no at equilibrium.

Answer 1

N2(g)+O2(g) <==> 2NO(g) Kc= .055 
Using the ICE Table:Kc =( [.01-2x]^2)/([x]*[x]) .055 = ( [.01-2x]^2)/([x]*[x]) Solve for x wherein x = .00445 
In conclusion, the equilibrium concentrations are as follows:[N2] = .00445M[O2] = .00445M[NO] = .0011M.

Answer 2

• [N₂] = 4.5 · 10⁻³ or 0.0045 M
• [O₂] = 4.5 · 10⁻³ or 0.0045 M
• [NO] = 0.001 M

Further explanation

Given:

• The equilibrium constant in terms of concentrations, Kc = 0.0055.
• The reaction mixture starts with only the product, [NO] = 0.0100.

Question:

Find the equilibrium concentrations of N₂, O₂, and NO at equilibrium.

The Process:

Consider the following equilibrium reaction.

$\boxed{ \ N_2 + O_2 \rightleftharpoons 2NO \ }$ The reaction is balanced.

We use the ICE table to find out all the concentrations of substances at equilibrium.

                              N₂    +   O₂       ⇄    2NO₂

Initial:                      -             -               0.0100

Change:                +x          +x                 -2x

Equilibrium:            x            x          (0.0100 - 2x)  

Let us write the equilibrium constant in terms of concentrations based on the reaction above.

$\boxed{ \ K_c = \frac{[NO_2]^2}{[N_2][O_2]} \ }$

Substitute all the data above into the equation.

$\boxed{ \ 0.055 = \frac{(0.0100 - 2x)^2}{(x)(x)} \ }$

$\boxed{ \ \frac{(0.0100 - 2x)^2}{x^2} = 0.055 \ }$

$\boxed{ \ 4x^2 - 4 \cdot 10^{-2}x + 10^{-4} = 0.055x^2 \ }$

$\boxed{ \ 3.945x^2 - 4 \cdot 10^{-2}x + 10^{-4} = 0 \ }$

We use the quadratic formula to get the roots.

$\boxed{ \ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ }$

After processing, the roots are as follows:

$\boxed{ \ x_1 \approx 4.5 \cdot 10^{-3} \ }$ and $\boxed{ \ x_2 \approx 5.7 \cdot 10^{-3} \ }$

Let us compare the two x values ​​to the equilibrium concentrations of NO₂.

• x₁ = 4.5 · 10⁻³ ⇒ 0.0100 - 2(4.5 · 10⁻³) = 0.001 (positive results, fulfilled)
• x₂ = 5.7 · 10⁻³ ⇒ 0.0100 - 2(5.7 · 10⁻³) = - 0.0014 (negative results, rejected)

Thus, the equilibrium concentrations of N₂, O₂, and NO at equilibrium are the following.

• [N₂] = 4.5 · 10⁻³ or 0.0045 M
• [O₂] = 4.5 · 10⁻³ or 0.0045 M
• [NO] = 0.001 M

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