Ethene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) reacts with halogens \(\left(\mathrm{X}_{2}\right)\) by the following reaction: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{X}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}(g) $$ The following figures represent the concentrations at equilibrium at the same temperature when \(\mathrm{X}_{2}\) is \(\mathrm{Cl}_{2}\) (green), \(\mathrm{Br}_{2}\) (brown), and \(\mathrm{I}_{2}\) (purple). List the equilibria from smallest to largest equilibrium constant. [Section 15.3\(]\)

Short Answer

Expert verified
The order of the equilibrium constants from smallest to largest is: \(K_\mathrm{I_2} < K_\mathrm{Br_2} < K_\mathrm{Cl_2}\).

Step by step solution

01

Understand the equilibrium constant (K) and its relation to reactants and products concentrations

The equilibrium constant (K) for a reaction is the ratio of the concentrations of the products to the reactants at equilibrium, raised to their stoichiometric coefficients. In this case, the reaction is: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{X}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}(g)$$ The equilibrium constant (K) for this reaction would be: $$K = \frac{[\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}]}{[\mathrm{C}_{2} \mathrm{H}_{4}][\mathrm{X}_{2}]}$$
02

Analyze the given figures (concentration plots)

The given figures show the concentrations of the reactants and products at equilibrium in each case. We are not given specific numerical values, but we can compare the relative heights of the bars to draw conclusions about the equilibrium constants. In each case, the higher the concentration of the halogen, the larger the equilibrium constant will be. Look at the relative heights of the product bars (\(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}\)) in comparison to the reactant bars (\(\mathrm{C}_{2} \mathrm{H}_{4}\) and \(\mathrm{X}_{2}\)) for each halogen: - For Cl2 (green): The product bar is taller than both reactant bars. - For Br2 (brown): The product bar is shorter than both reactant bars. - For I2 (purple): The product bar is shorter than the C2H4 bar, and about the same height as the I2 bar.
03

Determine the order of equilibrium constants

Based on the analysis in step 2, we can order the equilibrium constants as follows: 1. I2 (purple): The product concentration is less when compared to reactant concentrations, and as equilibrium constant is the ratio of product concentration over reactant concentrations, K is smallest in this case. 2. Br2 (brown): The product concentration is more than that of I2 while still being less than the reactant concentrations; so its equilibrium constant lies between that of I2 and Cl2 3. Cl2 (green): The product concentration is the highest with respect to reactant concentrations, thus its equilibrium constant is the largest. So, we can now list the equilibria from smallest to largest equilibrium constant: $$K_\mathrm{I_2} < K_\mathrm{Br_2} < K_\mathrm{Cl_2}$$

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Most popular questions from this chapter

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

Can the equilibrium constant ever be zero? Explain.

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (c) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

Consider the reaction \(\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\) If you start with \(25.0 \mathrm{~mL}\) of a \(0.905 \mathrm{M}\) solution of \(\mathrm{NaIO}_{4}\), and then dilute it with water to \(500.0 \mathrm{~mL}\), what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) at equilibrium?

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15\(), K_{p}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ is about \(1 \times 10^{-13}\). Assuming that the exhaust gas (total pressure 1 atm \()\) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2},\) and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction?

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