/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 The equilibrium constant \(K_{\m... [FREE SOLUTION] | 91Ó°ÊÓ

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The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) $$ has the value \(2.64 \times 10^{-3}\) at \(2300 . \mathrm{K}\). If a mixture of \(1.00 \mathrm{~mol} \mathrm{CO}\) and \(1.00 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) is allowed to come to equilibrium in a sealed, \(1.00-\mathrm{L}\) flask at \(2300 . \mathrm{K}\), (a) calculate the final concentrations of all four species: CO, \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2}\) (b) calculate the equilibrium concentrations after an additional 1.00 mol each of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) is added to the flask.

Short Answer

Expert verified
Equilibrium concentrations: CO, H2O = 0.9486 M; CO2, H2 = 0.0514 M. After adding reactants, recalculate new equilibrium similarly.

Step by step solution

01

Understanding the Initial Conditions

To begin, we know that the system initially contains 1.00 mol of CO and 1.00 mol of H2O in a 1 L flask, making their initial concentrations both 1.00 M. No CO2 or H2 is present initially, so their initial concentrations are 0 M.
02

Write the Equilibrium Expression

The equilibrium expression based on the reaction \( \text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{H}_2\text{(g)} \) is given by:\[ K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = 2.64 \times 10^{-3} \]
03

Set Up an ICE Table

An ICE (Initial, Change, Equilibrium) table will help track concentration changes:\[\begin{array}{c|c|c|c|c} & \text{CO} & \text{H}_2\text{O} & \text{CO}_2 & \text{H}_2 \hline\text{Initial (M)} & 1.0 & 1.0 & 0 & 0 \hline\text{Change (M)} & -x & -x & +x & +x \hline\text{Equilibrium (M)} & 1.0-x & 1.0-x & x & x \\end{array}\]
04

Substitute ICE Values into Equilibrium Expression

Substitute the equilibrium concentrations from the ICE table into the equilibrium expression:\[ 2.64 \times 10^{-3} = \frac{x^2}{(1.0-x)^2} \]
05

Solve the Quadratic Equation

Rearrange and solve the quadratic equation:\[ 2.64 \times 10^{-3} = \frac{x^2}{(1.0-2x+x^2)} \] Assuming \( x \) is small compared to 1, approximate \( (1-x)^2 \approx 1 \) which simplifies the equation to \( x^2 = 2.64 \times 10^{-3} \), giving \( x \approx \sqrt{2.64 \times 10^{-3}} = 0.0514 \).
06

Calculate Equilibrium Concentrations

Now calculate each equilibrium concentration:- \([\text{CO}] = 1.0 - 0.0514 = 0.9486 \text{ M}\)- \([\text{H}_2\text{O}] = 1.0 - 0.0514 = 0.9486 \text{ M}\)- \([\text{CO}_2] = 0.0514 \text{ M}\)- \([\text{H}_2] = 0.0514 \text{ M}\)
07

Re-calculate Equilibrium for Added Reactants

After adding 1.00 mol each of CO and H2O, initial concentrations become:- \([\text{CO}] = 1.9486 \text{ M}\)- \([\text{H}_2\text{O}] = 1.9486 \text{ M}\)Set up a new ICE table with these initial concentrations and solve a new equilibrium equation similarly, considering small \( x \), the same process can be repeated:- Results will be approximately: - \([\text{CO}] \approx 1.9486 - x\) - \([\text{H}_2\text{O}] \approx 1.9486 - x\) - \([\text{CO}_2] \approx x\) - \([\text{H}_2] \approx x\) with adjustments depending on the value of \( x \) after recalculation.

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Key Concepts

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

Equilibrium Constant
The concept of the equilibrium constant, denoted as \( K_c \), is crucial for understanding chemical equilibria. It provides a quantitative measure of the position of equilibrium in a reversible reaction. In simple terms, \( K_c \) reflects how much product and reactant are present when the system has reached equilibrium. If \( K_c \) is large, it suggests a lot of products at equilibrium; if small, more reactants remain.

For the reaction \( \mathrm{CO(g)} + \mathrm{H}_2\mathrm{O(g)} \rightleftharpoons \mathrm{CO}_2\mathrm{(g)} + \mathrm{H}_2\mathrm{(g)} \), the given equilibrium constant is \( 2.64 \times 10^{-3} \) at 2300 K. This value tells us that, at equilibrium, the system will have a much larger concentration of reactants compared to products. These calculations revolve around the balanced equation, as \( K_c \) is derived directly from it.

To calculate \( K_c \), use the expression:\[ K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} \]Here, brackets represent molar concentrations of the respective gases at equilibrium.
Reaction Quotient
The reaction quotient, \( Q \), is a tool used to determine the direction in which a reversible reaction will proceed to reach equilibrium. Like the equilibrium constant, \( Q \) is calculated using the same expression as \( K_c \), but with initial or non-equilibrium concentrations:

\[ Q = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} \]

By comparing \( Q \) to \( K_c \):
  • If \( Q < K_c \), the system will shift to the right, forming more products.
  • If \( Q > K_c \), the system will shift to the left, forming more reactants.
  • If \( Q = K_c \), the system is at equilibrium.

In our scenario, initially, \( [\text{CO}_2] = 0 \) M and \( [\text{H}_2] = 0 \) M, thus \( Q = 0 \), which is less than \( K_c \), indicating that the reaction will produce more CO2 and H2 to reach equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry used to predict the behavior of a system at equilibrium when it is subjected to an external change, such as concentration, temperature, or pressure. This principle states that if a system at equilibrium is disturbed, the system will adjust itself to counteract the disturbance and restore a new equilibrium.

In the exercise above, when additional 1.00 mol of CO and \( \text{H}_2\text{O} \) are added, the system is no longer in equilibrium. According to Le Chatelier's Principle, the system will respond by shifting the equilibrium to the right, producing more \( \text{CO}_2 \) and \( \text{H}_2 \), to reduce the effect of the increase in reactants.

This adjustment continues until a new equilibrium is established, where the concentrations align with the constant \( K_c \). This highlights how dynamic equilibria can adjust to changes, always striving to maintain balance.

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Most popular questions from this chapter

Chemists carried out a study of the high temperature reaction of sulfur dioxide with oxygen in which a sealed reactor initially contained \(0.0076-\mathrm{M} \mathrm{SO}_{2}, 0.0036-\mathrm{M} \mathrm{O}_{2}\), and no \(\mathrm{SO}_{3}\). After equilibrium was achieved, the \(\mathrm{SO}_{2}\) concentration decreased to \(0.0032 \mathrm{M}\). Calculate \(K_{\mathrm{c}}\) at this temperature for $$ 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) $$

The formation of hydrogen sulfide from the elements is exothermic. $$ \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{8} \mathrm{~S}_{8}(\mathrm{~s}) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=-20.6 \mathrm{~kJ} / \mathrm{mol} $$ Predict the effect of each of these changes on the position of the equilibrium; that is, state which way the equilibrium will shift (left, right, or no change) when each change is made in a constant-volume system. (a) Adding more sulfur (b) Adding more \(\mathrm{H}_{2}\) (c) Raising the temperature

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ has the value \(5.97 \times 10^{-2}\) at \(500 .{ }^{\circ} \mathrm{C}\). If \(1.00 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{gas}\) and \(1.00 \mathrm{~mol} \mathrm{H}_{2}\) gas are heated to \(500 .{ }^{\circ} \mathrm{C}\) in a \(10.00-\mathrm{L}\) sealed flask together with a catalyst, calculate the percentage of \(\mathrm{N}_{2}\) converted to \(\mathrm{NH}_{3}\). (Hint: Assume that only a very small fraction of the reactants is converted to products. Obtain an approximate answer and use it to obtain a more accurate result.)

A sealed 15.0 -L flask at \(300 . \mathrm{K}\) contains \(64.4 \mathrm{~g}\) of a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) in equilibrium. Calculate the total pressure in the flask. \(\left(\right.\) For \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) K_{\mathrm{P}}=\) 6.67 at \(300 . \mathrm{K} .)\)

Nitrosyl chloride, NOCl, decomposes to \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) at high temperatures. $$ 2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ Suppose you place \(2.00 \mathrm{~mol} \mathrm{NOCl}\) in a \(1.00-\mathrm{L}\) flask, seal it, and raise the temperature to \(462^{\circ} \mathrm{C}\). When equilibrium has been established, \(0.66 \mathrm{~mol} \mathrm{NO}\) is present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the decomposition reaction from these data.

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