Question

Here are some reactions and their equilibrium constants. (a) \(\begin{aligned} 2 \mathrm{CH}_{4}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) & \\ K_{\mathrm{c}}=9.5 \times 10^{-13} \end{aligned}\) (b) \(\mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) \(K_{\mathrm{c}}=3.6 \times 10^{20}\) (c) \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) \quad K_{\mathrm{c}}=2.0 \times 10^{9}\) Arrange these reactions in order of their increasing tendency to go to completion.

Short answer

The order of increasing tendency to go to completion is: (a), (c), (b).

Step-by-step solution

Interpret the Equilibrium Constant

Understand that the equilibrium constant (\(K_c\)) for a reaction at a given temperature indicates the extent to which the reaction proceeds to completion. A large value of the equilibrium constant indicates that the forward reaction is favored and the reaction has a high tendency to go to completion. Conversely, a small value of the equilibrium constant suggests that the reverse reaction is favored, and the reaction has a low tendency to go to completion.

Compare the Given Equilibrium Constants

Compile the equilibrium constants given for the reactions to analyze and compare their sizes. The larger the value of the equilibrium constant, the higher the tendency of the reaction to go to completion. Conversely, the smaller the constant, the lower the tendency to go to completion.

Arrange the Reactions

Order the reactions from the one with the smallest equilibrium constant to the one with the largest. This will be the order of increasing tendency to go to completion. (a): \(K_c = 9.5 \times 10^{-13}\), (b): \(K_c = 3.6 \times 10^{20}\), and (c): \(K_c = 2.0 \times 10^{9}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. It's vital to understand that equilibrium does not mean the reactants and products are present in equal amounts, but rather that their ratios are stable.

To visualize this, imagine a busy road where the number of cars entering and leaving is the same. The traffic (reactants and products) is constant, even though there could be many more cars on one side of the road than the other. It's the balance of traffic flow (reaction rates) that defines the equilibrium state, not the number of cars (concentrations).

When we say a reaction has reached equilibrium, it's akin to saying that a balance has been achieved where reactants and products are being formed and consumed at the same rate. It's important to note that this equilibrium can be reached from either direction of a reversible reaction, depending on the initial conditions and reactants' availability.

Reaction Completion Tendency

The tendency of a reaction to go to completion is an important concept when working with chemical reactions and can be inferred by examining the equilibrium constant. Completion in this context means that, ideally, all the reactants are converted into products.

To put it simply, just like some people have a strong inclination to finish any task they start while others may leave it midway, reactions also have varying tendencies to reach that 'final state' where reactants are fully converted to products.

Observing Reaction Tendencies

Consider a game where the chances of winning increase with the number of tries you make. Similarly, a reaction with a high equilibrium constant has more 'tries' or a higher probability of reaching completion than one with a low constant. This inclination to proceed towards products can be seen as the 'drive' or 'motivation' for a reaction to reach completion.

If we take the reaction \(2 CH_4(g) \rightleftharpoons C_2H_6(g) + H_2(g)\) with a small equilibrium constant \(K_c = 9.5 \times 10^{-13}\), we can say that this reaction has a low inclination to complete, preferring to stay mostly as reactants.

Equilibrium Constant Interpretation

Interpreting the equilibrium constant, \(K_c\), is like reading a book that tells us how likely a chemical reaction is to reach completion under a given set of conditions.

A higher value of \(K_c\) is like a strong recommendation in favor of the forward reaction, suggesting that you'll likely end up with more products. For instance, with \(K_c = 3.6 \times 10^{20}\) in the reaction \(CH_3OH(g) + H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)\), we can conclude that a significantly small amount of reactants remain as most have been converted to products.

In contrast, a lower value of \(K_c\), similar to a negative review, implies that the reverse reaction is the more favorable path, meaning that you'll have a substantial amount of reactants left with fewer products formed. This can be seen in the reaction given by \(K_c = 9.5 \times 10^{-13}\), indicating a preference for reactants to remain as they are.

Therefore, by interpreting the numerical value of the equilibrium constant, we can predict the dominant species in the chemical system at equilibrium - an invaluable tool for chemists in understanding and predicting the outcomes of chemical reactions.