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When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and \(1.50 \mathrm{~mol} \mathrm{H}_{2}\) are placed in a 3.00-L container at \(395^{\circ} \mathrm{C},\) the following reaction occurs: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) . \quad\) If \(\quad K_{c}=0.802\) what are the concentrations of each substance in the equilibrium mixture?

Short Answer

Expert verified
At equilibrium, the concentrations of the species are: \([\mathrm{CO}_{2}] = 0.277\,\mathrm{M}\), \([\mathrm{H}_{2}] = 0.277\,\mathrm{M}\), \([\mathrm{CO}] = 0.223\,\mathrm{M}\), and \([\mathrm{H}_{2}\mathrm{O}] = 0.223\,\mathrm{M}\).

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation is given as: \(\mathrm{CO}_{2}(g) + \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_{2}\mathrm{O}(g)\)
02

Set up an ICE table

An ICE (Initial, Change, Equilibrium) table helps track the changes in species concentrations throughout the reaction process. Initial concentrations: \(\frac{1.50\,\mathrm{mol}}{3.00\,\mathrm{L}} = 0.500\,\mathrm{M}\) for \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\) Initial concentrations of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\mathrm{O}\) are 0 M since they are not given in the problem statement. ``` CO2 H2 CO H2O Initial 0.500 0.500 0 0 Change -x -x +x +x Final 0.500-x 0.500-x x x ```
03

Write the equilibrium constant expression

\(K_c = \frac{[\mathrm{CO}] [\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{CO}_{2}] [\mathrm{H}_{2}]}\)
04

Substitute ICE table values into the equilibrium expression

We substitute the equilibrium concentrations from the ICE table into the \(K_c\) expression: \(0.802 = \frac{x^2}{(0.500-x)^2}\)
05

Solve for x

To solve for x, we first notice that the denominator and numerator have the same exponent. Thus, we can simplify the equation: \(0.802 = \frac{x}{0.500-x}\) Now, multiply both sides by \((0.500-x)\): \(0.802(0.500-x) = x\) Distribute the 0.802 on the left-hand side: \(0.401 - 0.802x = x\) Combine the x terms and solve for x: \(x = \frac{0.401}{1.802} \approx 0.223\,\mathrm{M}\)
06

Calculate the equilibrium concentrations

Now that we have the value of x, we can find the equilibrium concentrations of all species: \([\mathrm{CO}]_{eq} = [\mathrm{H}_{2}\mathrm{O}]_{eq} = x = 0.223\,\mathrm{M}\) \([\mathrm{CO}_{2}]_{eq} = [\mathrm{H}_{2}]_{eq} = 0.500 - x = 0.500 - 0.223 = 0.277\,\mathrm{M}\) So, at equilibrium, the concentrations of the species are: \([\mathrm{CO}_{2}] = 0.277\,\mathrm{M}\) \([\mathrm{H}_{2}] = 0.277\,\mathrm{M}\) \([\mathrm{CO}] = 0.223\,\mathrm{M}\) \([\mathrm{H}_{2}{\mathrm{O}}] = 0.223\,\mathrm{M}\)

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Key Concepts

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

ICE table
An ICE table is a valuable tool for understanding chemical reactions and their progress towards equilibrium. The term "ICE" stands for Initial, Change, and Equilibrium, reflecting the different stages of a reaction.
It's used to track the molar concentrations of reactants and products as the reaction progresses. Initially, we list what we know:
  • **Initial** concentrations: We begin by calculating the initial molarity of each reactant, using the formula: Molarity = moles/volume (in liters). For example, in our exercise, both the CO2 and H2 start with a concentration of 0.500 M.
  • **Change** in concentration: As the reaction proceeds, reactants are consumed, and products are formed. We represent these changes using a variable, often denoted as `x`. As CO2 and H2 react to form CO and H2O, they decrease by the amount `x`, while the products increase by `x`.
  • **Equilibrium** concentration: Finally, we express the equilibrium concentrations in terms of `x`. These are used in further calculations to find how much of each substance is present when the reaction reaches equilibrium.
This structured approach simplifies the often complex changes that occur in a reaction, making it easier to solve equilibrium problems.
equilibrium constant (Kc)
The equilibrium constant, or \(K_c\), is a key concept in chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation, at equilibrium.
In essence, \(K_c\) tells us how far a reaction proceeds before reaching equilibrium. If \(K_c > 1\), the products are favored at equilibrium, meaning more of the product is present than the reactant. Conversely, if \(K_c < 1\), the reactants are favored.
In mathematical terms, for a general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant expression is:
  • \(K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)
Here, the brackets denote molar concentrations.
In the exercise, the equilibrium constant \(K_c = 0.802\) indicates that, at equilibrium, there is a moderate amount of reaction towards products, but the reaction does not heavily favor them. This value is critical for determining concentrations of substances at equilibrium using the ICE table values.
reaction stoichiometry
Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It provides a blueprint of how much reactant is needed to form a particular amount of product.
In the exercise's reaction:\[\mathrm{CO}_{2}(g) + \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_{2}\mathrm{O}(g)\]each molecule of CO2 reacts with one molecule of H2 to produce one molecule of CO and one of H2O.
Stoichiometry is essential when filling in an ICE table:
  • We assign changes in concentration based on these stoichiometric coefficients. For example, for each mole of CO2 consumed, an equal amount of CO is produced.
  • This one-to-one relationship simplifies calculations, but in reactions with different coefficients, it’s crucial to correctly scale changes according to these ratios.
Reaction stoichiometry not only ensures that your ICE table is set up accurately, but it also provides the foundation for calculating equilibrium concentrations correctly. Understanding these stoichiometric relationships is key to mastering chemical equilibrium problems.

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Most popular questions from this chapter

Consider the hypothetical reaction \(\mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons\) \(2 \mathrm{C}(g),\) for which \(K_{c}=0.25\) at a certain temperature. A \(1.00-\mathrm{L}\) reaction vessel is loaded with \(1.00 \mathrm{~mol}\) of compound \(\mathrm{C}\), which is allowed to reach equilibrium. Let the variable \(x\) represent the number of \(\mathrm{mol} / \mathrm{L}\) of compound A present at equilibrium. (a) In terms of \(x,\) what are the equilibrium concentrations of compounds \(\mathrm{B}\) and \(\mathrm{C} ?\) (b) What limits must be placed on the value of \(x\) so that all concentrations are positive? (c) By putting the equilibrium concentrations (in terms of \(x\) ) into the equilibrium-constant expression, derive an equation that can be solved for \(x\). (d) The equation from part (c) is a cubic equation (one that has the form \(\left.a x^{3}+b x^{2}+c x+d=0\right)\). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of \(x\) that you specified in part (b). The point at which the cubic equation crosses the \(x\) -axis is the solution. (e) From the plot in part (d), estimate the equilibrium concentrations of \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\).

At \(1000 \mathrm{~K}, K_{p}=1.85\) for the reaction $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ (a) What is the value of \(K_{p}\) for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?(\mathbf{b})\) What is the value of \(K_{p}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) ? (c) What is the value of \(K_{c}\) for the reaction in part (b)?

In Section 11.5 we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the corresponding expression for \(K_{p} .\) (b) By using data in Appendix \(\mathrm{B}\), give the value of \(K_{p}\) for this reaction at \(30^{\circ} \mathrm{C} .(\mathrm{c})\) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

Nitric oxide (NO) reacts readily with chlorine gas as follows: $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g) $$ At \(700 \mathrm{~K}\) the equilibrium constant \(K_{p}\) for this reaction is \(0.26 .\) Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium. If not, state whether the mixture will need to produce more products or reactants to reach equilibrium. (a) \(P_{\mathrm{NO}}=0.15 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.31 \mathrm{~atm},\) and \(P_{\mathrm{NOCl}}=0.11 \mathrm{~atm} ;\) (b) \(P_{\mathrm{NO}}=0.12 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.10 \mathrm{~atm},\) and \(P_{\mathrm{NOCl}}=0.050 \mathrm{~atm} ;\) (c) \(P_{\mathrm{NO}}=0.15 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.20 \mathrm{~atm},\) and \(P_{\mathrm{NOCl}}=5.10 \times\) \(10^{-3}\) atm.

At \(100^{\circ} \mathrm{C}, K_{c}=0.078\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and \(\mathrm{SO}_{2}\) are \(0.108 \mathrm{M}\) and \(0.052 \mathrm{M}\), respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

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