Question

Given the equilibrium constant values $$\begin{array}{rl}{\mathrm{N}}_2(\mathrm{g})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})& \rightleftharpoons {\mathrm{N}}_2\mathrm{O}(\mathrm{g})K_{\mathrm{c}}=2.7\times {10}^{-18}\\ {\mathrm{N}}_2{\mathrm{O}}_4(\mathrm{g})& \rightleftharpoons 2{\mathrm{NO}}_2(\mathrm{g})K_{\mathrm{c}}=4.6\times {10}^{-3}\\ \frac{1}{2}{\mathrm{N}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})& \rightleftharpoons {\mathrm{NO}}_2(\mathrm{g})K_{\mathrm{c}}=4.1\times {10}^{-9}\end{array}$$ Determine a value of $K_{\mathrm{c}}$ for the reaction $$2{\mathrm{N}}_2\mathrm{O}(\mathrm{g})+3{\mathrm{O}}_2(\mathrm{g})\rightleftharpoons 2{\mathrm{N}}_2{\mathrm{O}}_4(\mathrm{g})$$

Short answer

The equilibrium constant ($K_c$) for the reaction 2 N2O(g) + 3 O2(g) ⇌ 2 N2O4(g) is 0.65

Step-by-step solution

Identify and Flip Equations

First, we identify which of the given equations will be useful to get the desired target equation. We can see that the second equation needs to be reversed so as to have N2O4 on the product side, and the first and third equations need to be doubled so as to get 2N2O and NO2, respectively, with the correct stoichiometric coefficients.

Find the New Equilibrium Constants

Calculate the new equilibrium constants. Recall the rules that apply to flipping or reversing a reaction (you take the reciprocal of the equilibrium constant, $K_c$) and for multiplying the entire equation by a factor (you raise the equilibrium constant to the power equal to the factor). For the first equation, we have $K_c^{\prime }=(2.7\times {10}^{-18}{)}^2$, for the second equation, $K_c^{\prime }=1/(4.6\times {10}^{-3})$, and for the third equation, $K_c^{\prime }=(4.1\times {10}^{-9}{)}^2$.

Compute $K_c^{\prime }C$'s and $K_c$ for the Desired Reaction

Determine the product of the altered $K_c^{\prime }$ values, which will be the desired equilibrium constant. Hence, $K_c=K_{c1}^{\prime }\times K_{c2}^{\prime }\times K_{c3}^{\prime }=(2.7\times {10}^{-18}{)}^2\times 1/(4.6\times {10}^{-3})\times (4.1\times {10}^{-9}{)}^2=0.65$

Key concepts

Equilibrium Constant

The equilibrium constant, often represented as $K_c$, quantifies the ratio of concentrations of products to reactants at equilibrium for a reversible reaction. Understanding the equilibrium constant helps us predict the direction of a reaction. If $K_c$ is significantly greater than 1, the reaction favors products, whereas if it's much less than 1, the reactants are favored.

Manipulating chemical equations affects the value of $K_c$. When an equation is reversed, we take the reciprocal of its $K_c$. If the coefficients in the equation are multiplied by a factor, $K_c$ is raised to the power of that factor. This allows us to modify given chemical equations to derive the desired equation's equilibrium constant. By mastering how to use and manipulate $K_c$, chemists can predict how changes in concentration affect the position of equilibrium in a reaction.

Reversible Reactions

Reversible reactions are fundamental in chemical equilibrium, where reactions can proceed in both forward and backward directions. At equilibrium, the rate of the forward reaction equals the rate of the backward reaction, resulting in a constant concentration of reactants and products.

For example, consider the reaction ${\mathrm{N}}_2+\frac{1}{2}{\mathrm{O}}_2\rightleftharpoons {\mathrm{N}}_2\mathrm{O}$. Here, the formation of ${\mathrm{N}}_2\mathrm{O}$ and its decomposition back into ${\mathrm{N}}_2$ and ${\mathrm{O}}_2$ can occur simultaneously. The equilibrium position depends on factors like temperature and initial concentrations, but at equilibrium, the amounts of reactants and products remain steady.

Understanding reversibility helps in predicting and controlling the outcome of reactions, crucial in industries where maximizing yield is essential. Recognizing reversible reactions and their equilibrium points ensures effective and controlled chemical processes.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It uses balanced chemical equations to determine the ratios of moles and masses needed or produced. This concept is essential in calculating how much of one substance is required to react with another completely without wastage.

To solve stoichiometric calculations, we first need a balanced chemical equation. For instance, in the reaction $2{\mathrm{N}}_2\mathrm{O}+3{\mathrm{O}}_2\rightleftharpoons 2{\mathrm{N}}_2{\mathrm{O}}_4$, for every 2 moles of ${\mathrm{N}}_2\mathrm{O}$, 3 moles of ${\mathrm{O}}_2$ are needed to produce 2 moles of ${\mathrm{N}}_2{\mathrm{O}}_4$. This mole ratio is critical in calculating reagent needs and assessing reaction yields.

With stoichiometry, scientists ensure that chemicals are used efficiently in experiments and industrial processes, minimizing costs and maximizing efficiency. It is vital for producing desired quantities of substances accurately and safely.