Question

For the reaction $2{\mathrm{NO}}_2(\mathrm{g})\rightleftharpoons 2\mathrm{NO}(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})$ $K_{\mathrm{c}}=1.8\times {10}^{-6}$ at ${184}^{\circ }\mathrm{C}.\mathrm{At}{184}^{\circ }\mathrm{C},$ the value of $K_{\mathrm{c}}$ for the reaction $\mathrm{NO}(\mathrm{g})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})\rightleftharpoons {\mathrm{NO}}_2(\mathrm{g})$ is (a) $0.9\times {10}^6$ (b) $7.5\times {10}^2$ (c) $5.6\times {10}^5$ (d) $2.8\times {10}^5$

Short answer

(b) $7.5\times {10}^2$

Step-by-step solution

Rearranging the given reaction

First, rearrange the provided reaction to match the reaction for which we need to find the equilibrium constant. The rearranged reaction will be $N{O}_{2(g)}\rightleftharpoons N{O}_{(g)}+1/2O_{2(g)}$

Determine the equilibrium constant for the rearranged reaction

In the rearranged reaction, notice that the stoichiometric coefficients are half of those in the original reaction. The equilibrium constant for the related reaction is the square root of the equilibrium constant for the original reaction. This is because the stoichiometric coefficients in the balanced chemical equation are exponents in the equation for the equilibrium constant. So, $K_{c(new)}=\sqrt{K_{c(original)}}=\sqrt{1.8\times {10}^{-6}}=1.34\times {10}^{-3}$.

Determine the equilibrium constant for the desired reaction

The desired reaction is the reverse of the rearranged reaction, so the equilibrium constant for the desired reaction is the reciprocal of the equilibrium constant for the rearranged reaction. Therefore, $K_{c(desired)}=1/K_{c(new)}=1/(1.34\times {10}^{-3})=0.75\times {10}^6$

Key concepts

Equilibrium Constant (Kc)

The equilibrium constant, denoted as $K_c$, is a crucial concept in understanding chemical equilibrium. It provides a numerical value that indicates the ratio of the concentrations of products to reactants at equilibrium. This ratio is specific to a given temperature and reaction. For the reaction $2{\mathrm{NO}}_2(g)\rightleftharpoons 2\mathrm{NO}(g)+{\mathrm{O}}_2(g)$, the equilibrium constant is $1.8\times {10}^{-6}$ at ${184}^{\circ }\mathrm{C}$.

The value of $K_c$ is influenced by the stoichiometry of the reaction; essentially, how many molecules of each reactant and product are involved. In essence, if you modify the stoichiometry, for example by dividing the equation coefficients by a number, you must also adjust the expression for $K_c$.

• For the original reaction $K_c$ is calculated from the concentrations raised to the powers of their coefficients in the balanced equation.
• If we change the coefficients, the relation between this and the equilibrium constant changes, requiring mathematical manipulation like square roots or reciprocals.

These computations showcase how flexible equilibrium constants are, depending on how we choose to express our chemical equation.

Reversible Reactions

Reversible reactions are a fascinating aspect of chemistry because they attain a state of dynamic equilibrium. When reactants form products and simultaneously, products revert into reactants, a reversible reaction is occurring. For example, the forward reaction $2{\mathrm{NO}}_2(g)\rightleftharpoons 2\mathrm{NO}(g)+{\mathrm{O}}_2(g)$ and its reverse are both happening.

In a closed system, reversible reactions will reach an equilibrium state where the rate of the forward reaction equals the rate of the backward reaction.

• This balancing act means that the concentrations of reactants and products remain constant over time at equilibrium.
• It's important to recognize that equilibrium does not mean that the reactants and products are equal but that their ratios remain constant as indicated by the $K_c$.

Understanding the dynamics of reversible reactions is essential since they form the foundation for calculating and working with equilibrium constants.

Stoichiometry in Chemical Reactions

Stoichiometry refers to the quantitative relationships between the amounts of reactants and products in a chemical reaction. It relates directly to the balanced chemical equation, ensuring that all atoms are conserved in the process. In equilibrium expressions, this concept takes on further importance.

For the reaction $2{\mathrm{NO}}_2(g)\rightleftharpoons 2\mathrm{NO}(g)+{\mathrm{O}}_2(g)$, the stoichiometric coefficients are essential for forming the correct equilibrium constant expression.

• For instance, if you alter the coefficients by halving them to get $N{O}_{\text{2(g)}}\rightleftharpoons N{O}_{\text{(g)}}+1/2O_{\text{2(g)}}$, you need to adjust the equilibrium constant by taking the square root, which alters $K_c$.
• In the case of reversing a reaction's direction, you take the reciprocal of $K_c$. This fundamental manipulation reflects stoichiometry's role in how we perceive chemical dynamics at equilibrium.

Mastering stoichiometry in reversible reactions is pivotal for understanding the math behind equilibrium constants and successfully manipulating chemical equations.