# Question #b826c

Mar 28, 2017

#### Explanation:

For starters, you must make sure that you're working with a balanced chemical equation

${\text{CO"_ ((g)) + "NO"_ (2(g)) rightleftharpoons "NO"_ ((g)) + "CO}}_{2 \left(g\right)}$

Now, the thing to remember about equilibrium reactions is that they are governed by Le Chaterlier's Principle, which states that a system at equilibrium will react to a stress placed on the current position of the equilibrium in such a way as to reduce that stress.

In other words, the system will react to any change in the position of the equilibrium by favoring the reaction that counteracts that change.

In your case, you must determine what change will cause the equilibrium to shift to the left, i.e. what change will favor the reverse reaction.

Right from the start, option A) looks like a match because if you increase the concentration of nitric oxide, the system will counteract this change by decreasing the concentration of nitric oxide.

Since the reverse reaction consumes nitric oxide and carbon dioxide and produces carbon monoxide and nitrogen dioxide, the equilibrium will shift to the left and the reverse reaction will be favored.

${\text{CO"_ ((g)) + "NO"_ (2(g)) rightleftharpoons overbrace("NO"_ ((g)))^(color(blue)("increase the amount of NO")) + "CO}}_{2 \left(g\right)}$

$\textcolor{w h i t e}{a a a a a a a a a a} \stackrel{\textcolor{g r e e n}{\leftarrow}}{\textcolor{w h i t e}{a a a a a a \textcolor{red}{\text{shift to the left}} a a a a a a a}}$

$\textcolor{w h i t e}{a}$

On to option B). If you remove carbon dioxide from the reaction vessel, the system will counteract this change by producing more carbon dioxide.

Consequently, the equilibrium will shift to the right and the forward reaction will be favored.

On to option C). This time, you're increasing the concentration of carbon monoxide. The system will counteract by consuming more carbon monoxide.

As a result, the equilibrium will shift to the right and the forward reaction will be favored.

Finally, let's look at option D). If you raise the total pressure inside the reaction vessel, the system will counteract the change by looking to decrease the pressure.

This can be done by decreasing the total number of moles of gas present in the vessel. So the system could counteract the increase in pressure by favoring the reaction that produces fewer moles of gas.

In your case, however, both the forward and the reverse reaction produce the same number of moles of gas, since you have

$\text{1 mole CO + 1 mole NO"_2 = "2 moles gas} \to$ on the reactants' side

$\text{1 mole NO + 1 mole CO"_2 = "2 moles gas} \to$ on the products' side

$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

SIDE NOTE This is why having balanced chemical equation to work with is so important. Notice that in the chemical equation you have listed, you have theree moles of gas on the products' side.

In this scenario, the system would counteract the increase in pressure by shifting to the left. However, that is not the case because the chemical equation is not balanced properly.

$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

Therefore, you can say that increasing the total pressure inside the reaction vessel will not affect the position of the equilibrium because both reactions produce the same number of moles of gas.

In other words, increasing the pressure will not cause the equilibrium to shift in any direction.