# Under a particular set of conditions the reaction comes to equilibrium where the partial pressures of N_2O_4 and NO_2 are found to be "0.6 atm" and "0.07" atm respectively. Determine K_p?

## The reaction is: ${N}_{2} {O}_{4} \left(g\right) r i g h t \le f t h a r p \infty n s 2 N {O}_{2} \left(g\right)$

Jan 23, 2018

${K}_{p} = 0.008$

#### Explanation:

You are working with the following equilibrium reaction

${\text{N"_ 2"O"_ (4(g)) rightleftharpoons color(red)(2)"NO}}_{2 \left(g\right)}$

For starters, the fact that, at equilibrium, the reaction vessel contains significantly more dinitrogen tetroxide than nitrogen dioxide--you can tell that this is the case because the equilibrium partial pressure of the reactant is higher than the equilibrium partial pressure of the product--tells you that you have

${K}_{p} < 1$

This means that at the temperature at which the reaction takes place, the reverse reaction is favored, which implies that the equilibrium will lie to the left.

In other words, if you start with dinitrogen tetroxide and no nitrogen dioxide, you should expect to find that most of the dinitrogen tetroxide will not react to produce nitrogen dioxide.

Similarly, if you start with nitrogen dioxide and no dinitrogen tetroxide, you should expect to find that most of the nitrogen dioxide will react to produce dinitrogen tetroxide.

The equilibrium constant that governs this reaction can be expressed using the equilibrium partial pressures of the two gases.

${K}_{p} = \left(\left({\text{NO"_ 2)^color(red)(2))/(("N"_ 2"O}}_{4}\right)\right)$

Plug in your values to find--I'll leave the expression of the equilibrium constant without added units!

${K}_{p} = {\left(0.07\right)}^{\textcolor{red}{2}} / \left(0.6\right) = 0.008$

The answer is rounded to one significant figure, the number of sig figs you have for your values.