Question

Giving the following reactions and associated equilibrium constants, what is the equilibrium constant for the following reaction?

Answer 1

Step 1 - Understanding the equations

Let's first note that the equation whose equilibrium constant is not known can be obtained from the previous equations. In order to do so, we have to sum the given equations, inverting them:

`\begin{gathered} SO_(3(g))\leftrightarrows SO_(2(g))+(1)/(2)O_(2(g)) \\ \\ NO_((g))+(1)/(2)O_(2(g))\leftrightarrows NO_(2(g)) \end{gathered}`

Note we have inverted the given equations. By summing these two equations, we will get the final one.

Step 2 - Manipulating the equilibrium constants

When we invert an equation, the equilibrium constant must also be inverted, like this:

`K_(inverted)=(1)/(K)`

Let's invert both equilibrium equations for reactions 1 and 2:

`\begin{gathered} K_{1,\text{ inv}}=(1)/(K_1)=(1)/(3.1)=0.32 \\ \\ K_{2,\imaginaryI\text{nv}}=(1)/(K_2)=(1)/(4.5)=0.22 \end{gathered}`

Step 3 - Finding the new equilibrium constant

The equilibrium constant of an equation that is the sum of other equations can be obtained by multiplying the equilibrium constants of the added equations.

Therefore, for this exercise, we have:

`K_(new)=K_(1,inv)* K_(2,inv)`

Substituting the constants of the equations we already know, we get:

`K_(new)=0.32*0.22=0.07`

Answer: K new = 0.07