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Consider the following system at equilibrium: $$ \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g) $$ Suppose the concentration of \(\mathrm{H}_{2} \mathrm{O}\) is increased. (a) In which direction does the reaction shift to reestablish equilibrium? (b) What happens to the concentrations of \(\mathrm{CH}_{4}, \mathrm{CO}_{2}\), and \(\mathrm{H}_{2}\) as the reaction shifts to reestablish equilibrium?

Short Answer

Expert verified
a) The reaction shifts to the left to reestablish equilibrium. b) The concentrations of \(\mathrm{CH}_{4}(g)\), \(\mathrm{CO}_{2}(g)\), and \(\mathrm{H}_{2}(g)\) decrease as the reaction shifts to reestablish equilibrium.

Step by step solution

01

Analyze the Effect of Increased \(\mathrm{H}_{2} \mathrm{O}\) Concentration

According to Le Chatelier’s principle, the reaction will shift in a direction that opposes the increase of \(\mathrm{H}_{2} \mathrm{O}\) concentration. Hence, it is expected that the system shifts to the left, therefor the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g)\) is favored.
02

Evaluate the Shift in Concentrations of Other Reactants and Products

With the shift of equilibrium towards the left, the system tries to consume some \(\mathrm{H}_{2} \mathrm{O}\) through the reverse reaction. This means that the concentrations of reactants like \(\mathrm{CH}_{4}(g)\) will decrease, while the concentrations of products like \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) will also decrease as they will be consumed in the reverse reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Understanding chemical equilibrium is crucial for anyone venturing into chemistry. Imagine a dance where both partners move in sync, sometimes stepping forward, sometimes backward, but overall, they remain in the same spot. That's the essence of chemical equilibrium - a state achieved when the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time.

For a chemical reaction like the one given \( \text{CH}_4(g) + 2 \text{H}_2O(g) \rightleftharpoons \text{CO}_2(g) + 4 \text{H}_2(g) \), equilibrium is reached when the amount of methane and water vapor being turned into carbon dioxide and hydrogen gas is matched by the reverse process. It's a delicate balance that, once established, can be maintained indefinitely, unless disrupted by changes in conditions such as temperature, pressure, or in our case, concentrations.

In practical terms, someone observing the reaction would notice that, after a certain point, the amounts of \( \text{CH}_4 \), \( \text{H}_2O \), \( \text{CO}_2 \), and \( \text{H}_2 \) would appear constant, though they are constantly reacting in a dynamic, but balanced, dance of chemistry.
Reaction Shift
When equilibrium is disturbed, the reaction doesn't just stand still; it responds and shifts to regain balance. This is where Le Chatelier's principle comes into play. Le Chatelier's principle suggests that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. It's like a cushion that adjusts its shape when you press down on it.

In our exercise, increasing the concentration of \( \text{H}_2O \) is like adding extra weight to one side of a balanced scale. The reaction counterbalances this by shifting the position of equilibrium to the left, favoring the reverse reaction to use up some of the excess \( \text{H}_2O \) and reducing its concentration. As the reaction adjusts, this shift can be observed by monitoring the changing concentrations of the involved substances.

This reactive dance, dictated by Le Chatelier’s principle, is not only fundamental in the lab but is also a cornerstone in industrial processes where maximizing product yield is key. By understanding and applying this principle, chemists can orchestrate the reaction conditions to shift the equilibrium in the desired direction.
Equilibrium Concentrations
Equilibrium concentrations refer to the specific amounts of reactants and products present when a chemical reaction is at equilibrium. These concentrations can be constant, but they are sensitive to changes in the reaction conditions. It's much like a recipe that's been perfected over time, but if one ingredient is altered, the outcome might change, and adjustments are needed to get it just right again.

Following the shift in equilibrium caused by the increased concentration of \( \text{H}_2O \), there will be a subsequent decrease in the concentrations of \( \text{CH}_4 \), \( \text{CO}_2 \), and \( \text{H}_2 \), as these components will react to form \( \text{H}_2O \) and offset its added amount. This alteration reflects the change in the equilibrium concentrations of both reactants and products as the system seeks to re-establish equilibrium according to Le Chatelier’s principle.

By monitoring these concentrations, chemists track the reaction's progress and understand the shifts in equilibrium. This knowledge is especially powerful when designing reactions for maximum efficiency, whether it's producing pharmaceuticals or synthesizing materials. Harmony in chemistry is not only about reaching equilibrium but also knowing how to nudge the system back into balance when it's tipped.

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Most popular questions from this chapter

Consider the following reaction: $$ \mathrm{CO}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{NO}(g) $$ At a specific temperature, the equilibrium concentrations were determined to be \([\mathrm{CO}]=0.033 M,\left[\mathrm{NO}_{2}\right]=0.021 M\), \(\left[\mathrm{CO}_{2}\right]=0.59 M\), and \([\mathrm{NO}]=0.59 \mathrm{M}\). (a) What is the value of the equilibrium constant? (b) Describe the position of the equilibrium.

What must you know about a reaction before you can predict the effect of a temperature change on the position of the equilibrium?

If the equilibrium constant for the reaction \(\mathrm{A} \rightleftharpoons \mathrm{B}\) is \(4.0\), what is the value of the equilibrium constant for the reaction \(\mathrm{B} \rightleftharpoons \mathrm{A}\) ?

Consider the following system at equilibrium: $$ \mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q) $$ Even small quantities of lead ion in drinking water can be toxic. Indicate whether or not the addition of each of the following would reduce the concentration of dissolved lead ion. Explain your reasoning for each. (a) \(\mathrm{PbI}_{2}\) (b) \(\mathrm{KI}\) (c) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)

(a) What is the mathematical relationship between the following two equilibrium constant expressions? $$ K_{\text {eq }}=\frac{[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{NO}_{2}\right]^{2}} \quad \text { and } \quad K_{\text {eq }}=\frac{\left[\mathrm{NO}_{2}\right]^{2}}{[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]} $$ (b) Write the balanced equations that correspond to each of these equilibrium constant expressions.

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