Question

chemical equilibrium problem N2 + O2 = 2NO equilibrium constant 0.10 mol/L at 2000 degrees Celsius find O2 concentration

Answer 1

To find the concentration of O2 at equilibrium for the reaction:

\[
N_2(g) + O_2(g) \rightleftharpoons 2NO(g)
\]

with an equilibrium constant \( K_c = 0.10 \, \text{mol/L} \) at 2000 °C, we start by writing the expression for the equilibrium constant \( K_c \):

\[
K_c = \frac{[NO]^2}{[N_2][O_2]}
\]

Let's assume that the initial concentrations of \( N_2 \) and \( O_2 \) are both \( x \, \text{mol/L} \) and that initially, there is no \( NO \). When the system reaches equilibrium, let \( y \) be the change in concentration of \( N_2 \) and \( O_2 \) that reacts:

- Change in concentration of \( N_2 \) and \( O_2 \): \(-y\)
- Change in concentration of \( NO \): \(+2y\)

The equilibrium concentrations will then be:

\[
[N_2] = x - y
\]
\[
[O_2] = x - y
\]
\[
[NO] = 2y
\]

Now substituting these equilibrium concentrations into the \( K_c \) expression:

\[
K_c = \frac{(2y)^2}{(x-y)(x-y)}
\]

Substituting \( K_c = 0.10 \):

\[
0.10 = \frac{4y^2}{(x-y)^2}
\]

From this equation, we can derive a relationship between \( y \) and \( x \). However, we need an initial concentration \( x \) for specific calculations.

To solve for \( [O_2] \), we can rearrange the equation:

\[
4y^2 = 0.10(x - y)^2
\]

For simplicity, let's pick a reasonable initial concentration (e.g., \( x = 1 \, \text{mol/L} \) as a hypothetical case). The equation becomes:

\[
4y^2 = 0.10(1 - y)^2
\]

Expanding and simplifying:

\[
4y^2 = 0.10(1 - 2y + y^2)
\]

\[
4y^2 = 0.10 - 0.20y + 0.10y^2
\]

Combining like terms leads to:

\[
4y^2 - 0.10y^2 + 0.20y - 0.10 = 0
\]

\[
3.90y^2 + 0.20y - 0.10 = 0
\]

Now applying the quadratic formula \( y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \):

Here, \( a = 3.90, b = 0.20, c = -0.10 \):

\[
y = \frac{-0.20 \pm \sqrt{(0.20)^2 - 4 \cdot (3.90) \cdot (-0.10)}}{2 \cdot (3.90)}
\]
\[
y = \frac{-0.20 \pm \sqrt{0.04 + 1.56}}{7.80}
\]
\[
y = \frac{-0.20 \pm \sqrt{1.60}}{7.80}
\]
\[
y = \frac{-0.20 \pm 1.2649}{7.80}
\]

Calculating both roots, we'll focus on the positive root because concentration cannot be negative:

\[
y = \frac{1.0649 - 0.20}{7.80} \approx 0.110 \, \text{mol/L}
\]

Now substituting back to find \( [O_2] \):

\[
[O_2] = x - y = 1.00 - 0.110 = 0.890 \, \text{mol/L}
\]

Thus, the equilibrium concentration of \( O_2 \) is approximately:

\[
\boxed{0.89 \, \text{mol/L}}
\]

Note: This is a hypothetical calculation based on an initial concentration you would define in a real scenario. Adjust \( x \) based on your initial conditions for more accurate results.