Animated Solution for Chemistry - Chemical Equilibrium: For the reaction, $2{\text{NO}}_2(g)\rightleftharpoons 2\text{NO}(g)+{\text{O}}_2(g)$ $[K_c=1.8\times 10^{-6}\text{ at }184^{\circ }\text{C},R=0.00831\text{ kJ}/(\text{mol K})]$ When $K_p$ and $K_c$ are compared at $184^{\circ }\text{C}$ it is found that

When dealing with gaseous reactions, we often encounter two types of equilibrium constants: $K_c$, which is based on molar concentrations, and $K_p$, which is based on partial pressures. But how do they relate to each other? Let's dive into this classic problem to find out.

We are given the decomposition reaction of nitrogen dioxide:

The first step in comparing $K_p$ and $K_c$ is to determine the change in the number of moles of gas, denoted as $\Delta n_g$. This is simply the sum of the stoichiometric coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants.

Looking at our balanced equation, we have $2$ moles of $\text{NO}$ and $1$ mole of ${\text{O}}_2$ on the product side, giving us a total of $3$ moles of gaseous products. On the reactant side, we have $2$ moles of ${\text{NO}}_2$.

Therefore, $\Delta n_g=(2+1)-2=1$.

The relationship between $K_p$ and $K_c$ is derived from the ideal gas law ($PV=nRT$) and is given by the formula:

Since we found that $\Delta n_g=1$, our equation simplifies beautifully to:

Now, we need to evaluate the term $RT$. The temperature is given as $184^{\circ }\text{C}$. Never forget to convert Celsius to Kelvin in thermodynamics!

The gas constant $R$ is given as $0.00831{\text{ kJ mol}}^{-1}{\text{K}}^{-1}$. Let's multiply them together:

Here is the crucial insight: we don't need to calculate the exact value of $K_p$. We just need to look at the multiplier. We have:

Because we are multiplying $K_c$ by a number strictly greater than $1$, the resulting value $K_p$ must be larger than $K_c$.

Final Answer: $K_p$ is greater than $K_c$.