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In the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K=31.4\) at \(588 \mathrm{K} .\) Equal masses of each reactant and product are brought together in a reaction vessel at \(588 \mathrm{K}\). (a) Can this mixture be at equilibrium? (b) If not, in which direction will a net change occur?

Short Answer

Expert verified
a) No, the mixture cannot be at equilibrium because the calculated equilibrium constant does not equal the given one. They will only equal when the reaction is at equilibrium. b) A net change will occur in the direction of forming more CO2 and H2.

Step by step solution

01

Understanding the Problem

In this exercise, a chemical reaction is given along with its equilibrium constant at a certain temperature. Equal masses of each reactant and product are taken in a reaction vessel at those same conditions: CO(g) + H2O(g) ↔ CO2(g) + H2(g), and K = 31.4 at 588K. It is then required to find out whether or not the mixture is in equilibrium, and if it is not, in what direction the reaction will proceed.
02

Determining the Initial State

Since it is mentioned that equal masses of each reactant and product are present initially, we can assume that their concentrations are also equal as chemical concentrations depend on masses and not on the nature of the substances. Let's assign this common concentration as 'a'.
03

Determining the Equilibrium Constant

Next, we can write our equilibrium constant expression as per the law of mass action. For the reaction CO(g) + H2O(g) ↔ CO2(g) + H2(g), the equilibrium constant K= [CO2][H2] / [CO][H2O]. As per the given condition, we have K = (a^2) / (a^2).
04

Comparing Given and Calculated Equilibrium Constants

The calculated equilibrium constant should equal the value that is given, which is 31.4. However, our calculated K is equal to 1, because (a^2) / (a^2) simplifies directly to 1.
05

Conclusion

Based on the comparison, we see that the given chemical system is not in equilibrium, because the calculated K does not match the given K. Therefore, the reaction needs to proceed in the direction that makes the two equal.
06

Direction of Reaction

Since calculated K (which equals 1) is less than the given K (which is 31.4), the reaction needs to proceed in the direction that increases the value of K. This is the direction that forms more products, which is the forward direction. So, the answer to (b) is that a net change will occur in the direction of forming more CO2 and H2.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant is a crucial concept in chemical reactions. It is denoted as \( K \) and serves as a measure of the ratio of the concentrations of products to reactants at equilibrium for a particular reaction. This constant is temperature-dependent and unique for every reaction.
For the given reaction \(\mathrm{CO(g)} + \mathrm{H}_2\mathrm{O(g)} \rightleftharpoons \mathrm{CO}_2\mathrm{(g)} + \mathrm{H}_2\mathrm{(g)}\), the equilibrium constant \( K \) is 31.4 at 588 K. The expression for the equilibrium constant based on the concentrations \([\mathrm{CO}_2]\), \([\mathrm{H}_2]\), \([\mathrm{CO}]\), and \([\mathrm{H}_2\mathrm{O}]\) would be:
  • \( K = \frac{[\mathrm{CO}_2 ][\mathrm{H}_2]}{[\mathrm{CO}][\mathrm{H}_2\mathrm{O}]} \)
The value of \( K \) provides insight into the balance between reactants and products at equilibrium. A large \( K \) (greater than 1) usually indicates a reaction heavily favoring product formation. Conversely, a small \( K \) (less than 1) suggests that reactants are more prevalent at equilibrium.
Reaction Direction
The direction in which a reaction proceeds can be determined by comparing the current state of the system with its state at equilibrium. If a reaction mixture is not at equilibrium, changes will occur so that equilibrium is eventually reached.

In the exercise, we set out to determine the direction of change for our reaction:
  • If \( Q < K \): The reaction will proceed in the forward direction (towards products) because the reaction quotient \( Q \) suggests a lower concentration of products than expected at equilibrium.
  • If \( Q > K \): The reaction will proceed in the reverse direction (towards reactants).
An initial calculation showed that \( Q \), the reaction quotient, is 1 when equal concentrations of reactants and products are assumed. Since this \( Q \) is less than the given \( K \) of 31.4, the reaction will move forward, increasing the concentration of \( \mathrm{CO}_2 \) and \( \mathrm{H}_2 \) to move towards equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that describes how a system at equilibrium responds to disturbances or changes in conditions like concentration, temperature, or pressure.

This principle allows us to predict how the system will shift to restore equilibrium. For instance:
  • **Pressure Change**: If the pressure is increased, the system will favor the side with fewer gas molecules.
  • **Concentration Change**: Adding more reactants will shift equilibrium towards the products, while adding more products will shift it towards reactants.
  • **Temperature Change**: An increase in temperature can favor the endothermic direction, while a decrease favors the exothermic direction.
In the current scenario, without any external changes, the observed imbalance in \( K \) and \( Q \) naturally spurs a shift to the forward reaction. According to Le Chatelier's Principle, since the system is not at equilibrium, it will adjust by forming more \( \mathrm{CO}_2 \) and \( \mathrm{H}_2 \), decreasing reactant concentrations and increasing product concentrations until equilibrium is achieved, aligning \( Q \) with \( K \).

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Most popular questions from this chapter

Based on these descriptions, write a balanced equation and the corresponding \(K_{c}\) expression for each reversible reaction. (a) Carbonyl fluoride, \(\mathrm{COF}_{2}(\mathrm{g}),\) decomposes into gaseous carbon dioxide and gaseous carbon tetrafluoride. (b) Copper metal displaces silver(I) ion from aqueous solution, producing silver metal and an aqueous solution of copper(II) ion. (c) Peroxodisulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\), oxidizes iron(II) ion to iron(III) ion in aqueous solution and is itself reduced to sulfate ion.

Lead metal is added to \(0.100 \mathrm{M} \mathrm{Cr}^{3+}(\mathrm{aq}) .\) What are \(\left[\mathrm{Pb}^{2+}\right],\left[\mathrm{Cr}^{2+}\right],\) and \(\left[\mathrm{Cr}^{3+}\right]\) when equilibrium is established in the reaction? $$\begin{aligned} \mathrm{Pb}(\mathrm{s})+2 \mathrm{Cr}^{3+}(\mathrm{aq}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cr}^{2+}(\mathrm{aq}) & \\ K_{\mathrm{c}}=3.2 \times 10^{-10} & \end{aligned}$$

A sample of pure \(\mathrm{PCl}_{5}(\mathrm{g})\) is introduced into an evacuated flask and allowed to dissociate. $$ \mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ If the fraction of \(\mathrm{PCl}_{5}\) molecules that dissociate is denoted by \(\alpha,\) and if the total gas pressure is \(P\) show that $$ K_{\mathrm{p}}=\frac{\alpha^{2} P}{1-\alpha^{2}} $$

Briefly describe each of the following ideas or phenomena: (a) dynamic equilibrium; (b) direction of a net chemical change; (c) Le Châtelier's principle; (d) effect of a catalyst on equilibrium.

Starting with \(\mathrm{SO}_{3}(\mathrm{g})\) at \(1.00 \mathrm{atm},\) what will be the total pressure when equilibrium is reached in the following reaction at \(700 \mathrm{K} ?\) \(2 \mathrm{SO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.6 \times 10^{-5}\)

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