# How to find partial pressures from a given equilibrium constant Kp?

I answered wrong on my quiz the following question on finding an equilibrium constant:

For the following reaction, at $$1~\mathrm{bar}$$ and $$300~\mathrm{K}$$, partial pressures of $$\ce{NO2}$$ and $$\ce{N2O4}$$ are both $$0.5~\mathrm{bar}$$. If pressure is increased to $$2~\mathrm{bar}$$ (this is the final pressure), and temperature is increased to $$400~\mathrm{K}$$, calculate the final composition of the system. $$\Delta H$$ for this reaction is $$57.2~\mathrm{kJ/mol}$$; assume it is independent of temperature.

$$\ce{N2O4 (g) <=> 2NO2 (g)}$$

My attempt:

I first found $$K_\textrm{p}$$:

$$K_p (300~\mathrm{K}) = \frac{P^2(\ce{NO2})}{P(\ce{N2O4})} = (0.5^2)/0.5 = 0.5$$

Then I used

$$\ln\left(\frac{K_p(400~\mathrm{K})}{K_p(300~\mathrm{K})}\right) = - \frac{\Delta H}{\mathcal{R}} \left( \frac{1}{400~\mathrm{K}} - \frac{1}{300~\mathrm{K}}\right)$$

I solved for $$K_\textrm{p}$$ at $$\pu{400 K} = 92695.44$$

This is where everything went wrong. I can not figure out how to get from $$K_p$$ to partial pressures. All I got to is the following line:

$$P_\text{total} = 2P(\ce{NO2}) + P(\ce{N2O4})$$, since $$n(\ce{NO2})$$ to $$n(\ce{N2O4})$$ ratio is $$2:1$$. I assumed they exert pressure proportionally.

Can some one guide me in the right direction?

1. $$P(\ce{N2O4(g)}) = 0.0252~\mathrm{bar}$$; $$P(\ce{NO2(g)}) = 1.9748~\mathrm{bar}$$
2. $$P(\ce{N2O4(g)}) = 0.039~\mathrm{bar}$$; $$P(\ce{NO2(g)}) = 1.961~\mathrm{bar}$$
3. $$P(\ce{N2O4(g)}) = 1.45~\mathrm{bar}$$; $$P(\ce{NO2(g)}) = 0.098~\mathrm{bar}$$
4. $$P(\ce{N2O4(g)}) = 1~\mathrm{bar}$$; $$P(\ce{NO2(g)}) = 1~\mathrm{bar}$$
• I just calculated the answer by doing some answer to question maping. essentially, I solved for ln(Kp at 400K) on following equation ln(Kp at 400k / Kp at 300k) = - (delta H / R) ( 1/ 400K - 1/300K) Than substitute all the values from multiple choices. current answer should be 1. But I still would like to know how to actually solve it. Mar 24, 2015 at 5:57
• Please verify the given numbers! Mar 24, 2015 at 10:02
• At the initial conditions, the mole fractions of $\ce{N2O4}$ and $\ce{NO2}$ are both 0.5. Take as a basis 1 mole of gas to start with at the initial conditions. Let $x$ represent the number of moles of $\ce{N2O4}$ that react to form $\ce{NO2}$. In terms of $x$, what are the new numbers of moles of these species? What are the new mole fractions? What are the new partial pressures? Mar 24, 2015 at 15:36
• Please note that in reality $K_p=0.14 bar$ for the reaction in your problem, not 0.5. This of course should not matter assuming this is considered only an exercise. It does matter however whether you provided the right answers to check the validity. May 7, 2021 at 12:24

Let:

$$A$$ represent $$N_2O_4$$

$$C$$ represent $$NO_2$$

Then the reaction becomes:

$$\ce{A <=> 2C}$$

$$K_{p1}=0.5 bar$$ (@ 300K)

$$K_{p2}=154.76bar$$ (@ 400K)

Let:

$$P_{A2}$$ = final equilibrium partial pressure of A

$$P_{C2}$$ = final equilibrium partial pressure of C

$$P_2$$ = Final total equilibrium pressure

Then:

$$P_2=P_{A2}+P_{C2}=2$$

We can then put $$P_{A2}$$ in terms of $$P_{C2}$$:

$$P_{A2}=2-P_{C2}$$

We then substitute into the expression of $$K_{p2}$$:

$$K_{p2}=\frac{P_{C2}^2}{P_{A2}}=\frac{P_{C2}^2}{2-P_{C2}}=154.76$$

Solving for $$P_{C2}$$, and then for $$P_{A2}$$:

$$P_{C2}=1.9748$$ bar

$$P_{A2}=0.252$$ bar

So the final answer would be (1)

$$\require{cancel}$$ This is a partial solution to this question which involves a bit of using the state of the answers to answer the question.

As OP has suggested, the first step in this is calculating $$K_\mathrm{p}(\pu{400 K})$$, for this we need $$K_\mathrm p (\pu{300 K})$$

$$K_\mathrm p(\pu{300 K}) = \frac{P(\ce{NO2})^2}{P\ce{(N2O4)}} = 0.5$$

$$\ln\left(\frac{K_\mathrm p(\pu{300 K})}{K_\mathrm p(\pu{400 K})} \right) = -\frac{\Delta H}{R}\left(\frac{1}{\pu{400 K}} - \frac{1}{\pu{300 K}}\right)$$

Upto this point the answer is correct. However, there is a mistake in the calculation, taking $$R = \pu{8.314 J K-1 mol-1}$$, and $$\Delta H = \pu{57200 J mol-1}$$, we get:

$$\ln\left(\frac{K_\mathrm p(\pu{300 K})}{K_\mathrm p(\pu{400 K})} \right) = \frac{572\cancel{00} \times \cancel{100}}{8.314 \times \cancelto{3}{300} \times \cancelto{4}{400}}$$

Upon further solving, we get,

$$\ln\left(\frac{K_\mathrm p(\pu{300 K})}{K_\mathrm p(\pu{400 K})} \right) = -5.7333$$

Therefore, $$K_\mathrm p(\pu{400 K})$$ can be found to be:

\begin{align} K_\mathrm p(\pu{400 K}) &= K_\mathrm p(\pu{300 K})\cdot e^{5.7333} \\ & = \pu{154.49 bar} \end{align}

Now, the intended (and lengthy solution) intends for you to find the change in moles of $$\ce{NO2}$$ and $$\ce{N2O4}$$ and continue the solution.

As Chet Miller very well pointed out,

At the initial conditions, the mole fractions of $$\ce{N2O4}$$ and $$\ce{NO2}$$ are both 0.5. Take as a basis 1 mole of gas to start with at the initial conditions. Let x represent the number of moles of $$\ce{N2O4}$$ that react to form $$\ce{NO2}$$. In terms of x, what are the new numbers of moles of these species? What are the new mole fractions? What are the new partial pressures?

(emphasis mine)

Answer those two questions, find the value of x, and find the number of moles of each compound which would give you your answer about the intermediate steps that are involved in this solution. However, it is lengthy and arduous to show those steps.

A much shorter (and possibly intended) method involves using the state of the options provided. You firstly need to notice that there are only two possible options that provide an answer to this question. They are (A) and (B). Why? This is because the value of $$K_p$$ is greater than one which means that $$P(\ce{NO2}) > P(\ce{N2O4})$$. Now substituting these values, we get:

$$K_p (\textbf{A}) = \pu{154.76 bar}$$ $$K_p (\textbf{B}) = \pu{98.603 bar}$$

Within error of round off, the correct answer would be (A).