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: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) .\) If \(K_{c}=0.802\) what are the concentrations of each substance in the equilibrium mixture?

Short Answer

Expert verified
The equilibrium concentrations of the substances in the reaction are approximately: CO₂: 0.052 M, H₂: 0.052 M, CO: 0.448 M, and H₂O: 0.448 M.

Step by step solution

01

Write down the balanced equation

First, write the balanced chemical equation for the reaction: \[ CO_2(g) + H_2(g) \rightleftharpoons CO(g) + H_2O(g) \]
02

Determine initial concentrations

Next, determine the initial concentrations of the reactants and products using the moles and volume of the container: Initial concentration of CO₂ = \(\frac{1.50\: mol}{3.00\: L} = 0.500\: M\) Initial concentration of H₂ = \(\frac{1.50\: mol}{3.00\: L} = 0.500\: M\) Initially, no reaction has occurred, so concentrations of CO and H₂O are both 0 M.
03

Set up the ICE table

Now, we'll set up an ICE (Initial, Change, Equilibrium) table for the reaction: | | CO₂ | H₂ | CO | H₂O | |-----|-----|-----|-----|-----| | I | 0.5 | 0.5 | 0 | 0 | | C | -x | -x | +x | +x | | E |0.5-x|0.5-x| x | x | Where 'I' represents the initial concentrations, 'C' represents the change in concentrations as the reaction proceeds, and 'E' represents the equilibrium concentrations.
04

Write the expression for K_c

Write the expression for K_c using the equilibrium concentrations of species: \[ K_c = \frac{[CO][H_2O]}{[CO_2][H_2]} \] Given that \(K_c = 0.802\), we can substitute the equilibrium concentrations from the ICE table: \[ 0.802 = \frac{(x)(x)}{(0.5-x)(0.5-x)} \]
05

Solve the equation for x

Now, we'll solve the equation for x. It's a quadratic equation, but since the value of K_c is much less than 1, we can assume that the change in concentrations is small and x is much smaller than 0.5: \[ 0.802 = \frac{x^2}{(0.5)(0.5)} \] Solve for x: \[ x^2 = 0.802(0.25) \] \[ x = \sqrt{0.2005} \approx 0.448\: M \]
06

Determine the equilibrium concentrations

Use the value of x to find the equilibrium concentrations of all species: Equilibrium concentration of CO₂ = 0.5 - x ≈ \(0.5 - 0.448 \approx 0.052\: M\) Equilibrium concentration of H₂ = 0.5 - x ≈ \(0.5 - 0.448 \approx 0.052\: M\) Equilibrium concentration of CO = x ≈ \(0.448\: M\) Equilibrium concentration of H₂O = x ≈ \(0.448\: M\) So, the equilibrium concentrations are approximately: CO₂: 0.052 M H₂: 0.052 M CO: 0.448 M H₂O: 0.448 M

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Key Concepts

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

Equilibrium Constant (Kc)
Understanding the concept of an equilibrium constant, denoted as \( K_c \), is vital in analyzing chemical reactions. It is a numerical value that reflects the ratio of the concentrations of products to the concentrations of reactants at equilibrium, each raised to the power of their respective coefficients in the balanced equation.

For a generalized reaction where \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K_c \) is given by the expression \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \). Here, the square brackets denote concentration in molarity (\( M \)) at equilibrium. The constant itself is unitless and provides insight into the position of the equilibrium; a high \( K_c \) indicates a reaction favoring products, while a low \( K_c \) indicates a reaction favoring reactants. In the given problem, the \( K_c \) is 0.802, suggesting a close balance between reactants and products.
ICE Table
An ICE (Initial, Change, Equilibrium) table is a structured approach to determining the equilibrium concentrations of reactants and products in a chemical reaction. It outlines the initial concentrations, the changes that occur as the system reaches equilibrium, and the final equilibrium concentrations.

The 'I' column lists the initial concentrations before any reaction has occurred. The 'C' column denotes the change in concentration in terms of an unknown variable (often 'x'), which represents the amount of reactant consumed and product formed. The 'E' column then lists expressions for the equilibrium concentrations, which are combinations of the initial concentrations and the changes. Using the ICE table is a methodical way to visualize and calculate the shifts in a reacting system and is crucial for solving equilibrium problems, as demonstrated in the provided solution.
Equilibrium Concentrations
Equilibrium concentrations are the concentrations of each species (reactants and products) in a reaction mixture when the forward and reverse reaction rates are equal, i.e., the system is in a state of dynamic balance. Determining these values is essential to understanding the extent of a reaction and its yield.

Estimation of equilibrium concentrations often involves setting up an equation based on the equilibrium constant expression, substituting the concentrations as defined in the ICE table, and solving for the unknown variable. Once the variable is found, it is substituted back into the equilibrium expressions from the ICE table to find the concentrations of all substances present at equilibrium. Knowing these amounts can provide a detailed picture of the reaction's behavior at a given temperature.
Reaction Quotient
The reaction quotient, \( Q \), serves as a gauge for the progress of a reaction, helping one to determine whether a system is at equilibrium and, if not, in which direction it needs to shift to reach equilibrium. \( Q \) is calculated using the same formula as the equilibrium constant \( K_c \), but with the current concentrations instead of the equilibrium concentrations.

If \( Q \) is less than \( K_c \), the forward reaction is favored, and the system will shift towards the products to reach equilibrium. Conversely, if \( Q \) is greater than \( K_c \), the reverse reaction is favored, and the system will shift towards the reactants. At equilibrium, \( Q \) equals \( K_c \), indicating that the forward and reverse reactions are occurring at the same rate.

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Most popular questions from this chapter

Gaseous hydrogen iodide is placed in a closed container at \(425^{\circ} \mathrm{C}\) , where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) .\) At equilibrium it is found that \([\mathrm{HI}]=3.53 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2}\right]=4.79 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{I}_{2}\right]=4.79 \times 10^{-4} \mathrm{M} .\) What is the value of \(K_{c}\) at this temperature?

At \(700 \mathrm{K},\) the equilibrium constant for the reaction $$\operatorname{CCI}_{4}(g) \Longrightarrow \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ is \(K_{p}=0.76 .\) A flask is charged with 2.00 atm of \(\mathrm{CCl}_{4}\) ,which then reaches equilibrium at 700 \(\mathrm{K}\) . (a) What fraction of the CCl\(_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?(\mathbf{b})\) what are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

Mercury(l) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium- constant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

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 -L reaction vessel is loaded with 1.00 mol of compound \(C,\) which is allowed to reach equilibrium. Let the variable \(x\) represent the number of mol/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 \(x .(\mathbf{d})\) The equation from part (c) is a cubic equation (one that has the form \(a x^{3}+b x^{2}+c x+d=0 )\) . 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 \(A, B,\) and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

Consider the reaction $$\begin{array}{rl}{4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g)} & {\rightleftharpoons} \\ {4} & {\mathrm{NNO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g), \Delta H=-904.4 \mathrm{kJ}}\end{array}$$ Does each of the following increase, decrease, or leave unchanged the yield of \(\mathrm{NO}\) at equilibrium? (a) increase \(\left[\mathrm{NH}_{3}\right] ;(\mathbf{b})\) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\mathbf{c})\) decrease \(\left[\mathrm{O}_{2}\right] ;(\mathbf{d})\) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.

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