Problem: Even at high T, the formation of NO is not favored: N2(g) + O2(g) ⇌ 2NO(g)         Kc = 4.10×10−4 at 2000°C What is [NO] when a mixture of 0.20 mol of N 2(g) and 0.15 mol of O2(g) reach equilibrium in a 1.0-L container at 2000°C?

🤓 Based on our data, we think this question is relevant for Professor Wink's class at UIC.

FREE Expert Solution

N2(g) + O2(g) ⇌ 2NO(g)         Kc = 4.10×10−4 at 2000°C

Step 1: Calculate the initial concentrations.

[N2] = 0.20 M

[O2] = 0.15 M

Step 2: Construct an ICE chart for the equilibrium reaction.

Step 3: Calculate the change (x) in the reaction using the equilibrium constant.

N2(g) + O2(g) ⇌ 2NO(g)         Kc = 4.10×10−4

${\mathbf{K}}_{\mathbf{c}}\mathbf{=}\frac{{\mathbf{\left[}\mathbf{NO}\mathbf{\right]}}^{\mathbf{2}}}{\mathbf{\left[}{\mathbf{N}}_{\mathbf{2}}\mathbf{\right]}\mathbf{\left[}{\mathbf{O}}_{\mathbf{2}}\mathbf{\right]}}\phantom{\rule{0ex}{0ex}}\mathbf{4}\mathbf{.}\mathbf{10}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{=}\frac{{\mathbf{\left(}\mathbf{2}\mathbf{x}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{-}\mathbf{x}\mathbf{\right)}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{-}\mathbf{x}\mathbf{\right)}}\phantom{\rule{0ex}{0ex}}\mathbf{4}\mathbf{.}\mathbf{10}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{03}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{4}{\mathbf{x}}^{\mathbf{2}}}{\overline{)\mathbf{0}\mathbf{.}\mathbf{03}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}}\mathbf{\left(}\overline{)\mathbf{0}\mathbf{.}\mathbf{03}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{435}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\right)}\mathbf{x}\mathbf{+}\mathbf{\left(}\mathbf{4}\mathbf{.}\mathbf{10}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\right)}{\mathbf{x}}^{\mathbf{2}}\mathbf{=}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\overline{)\mathbf{0}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{99959}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{435}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\right)}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}}$

Now, we will use the quadratic formula to solve for x. The quadratic formula is:

$\overline{){\mathbf{x}}{\mathbf{=}}\frac{\mathbf{-}\mathbf{b}\mathbf{±}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}}$

Solving for x:

$\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{435}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\right)}\mathbf{±}\sqrt{\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{435}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{\right)}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{\left(}\mathbf{3}\mathbf{.}\mathbf{99959}\mathbf{\right)}\mathbf{\left(}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{\right)}}}{\mathbf{2}\mathbf{\left(}\mathbf{3}\mathbf{.}\mathbf{99959}\mathbf{\right)}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{435}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\right)}\mathbf{±}\sqrt{\mathbf{2}\mathbf{.}\mathbf{059225}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{9678}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}}{\mathbf{7}\mathbf{.}\mathbf{99918}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{435}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\right)}\mathbf{±}\sqrt{\mathbf{1}\mathbf{.}\mathbf{968}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}}{\mathbf{7}\mathbf{.}\mathbf{99918}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{435}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\right)}\mathbf{±}\mathbf{0}\mathbf{.}\mathbf{014}}{\mathbf{7}\mathbf{.}\mathbf{99918}}$

The calculation then splits into two ways:

Problem Details

Even at high T, the formation of NO is not favored:

N2(g) + O2(g) ⇌ 2NO(g)         Kc = 4.10×10−4 at 2000°C

What is [NO] when a mixture of 0.20 mol of N 2(g) and 0.15 mol of O2(g) reach equilibrium in a 1.0-L container at 2000°C?