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

Short Answer

Expert verified
The equilibrium concentrations are CO2: 0.283 M, H2: 0.283 M, CO: 0.217 M, and H2O: 0.217 M.

Step by step solution

01

Write the equilibrium expression for the reaction

For the reaction: \[CO_{2}(g) + H_{2}(g) \rightleftharpoons CO(g) + H_{2}O(g)\] The equilibrium constant expression can be written as: \[K_c = \frac{[CO][H_2O]}{[CO_2][H_2]}\]
02

Calculate the initial concentrations of each substance

We are given that 1.50 moles of CO2 and 1.50 moles of H2 are placed in a 3.00 L container. Therefore, the initial concentrations are: \[ [CO_2]_0 = \frac{1.50\, \text{mol}}{3.00\, \text{L}} = 0.50\, \text{M} \] \[ [H_2]_0 = \frac{1.50\, \text{mol}}{3.00\, \text{L}} = 0.50\, \text{M} \] Initially, there is no CO and H2O in the container since the reaction has not occurred yet, so: \[ [CO]_0 = [H_2O]_0 = 0 \, \text{M} \]
03

Set up an ICE table

Now, we can set up an ICE table to keep track of the change in concentrations as the reaction occurs and reaches equilibrium: ``` | CO2 H2 CO H2O ---------------------------------------------- Initial | 0.50 M 0.50 M 0 M 0 M Change | -x -x +x +x Equilibrium|0.50-x 0.50-x x x ```
04

Use the equilibrium expression to solve for the unknown equilibrium concentrations

We can now substitute the equilibrium concentrations from the ICE table into the equilibrium constant expression and solve for x: \[K_c = \frac{[CO][H_2O]}{[CO_2][H_2]} = \frac{x \cdot x}{(0.50 - x)(0.50 - x)} = 0.802\] Solving this quadratic equation for x: \[ x^2 = 0.802 (0.50 - x)^2 \] We will find that x has only one physically meaningful solution: \[x = 0.217\, \text{M}\] Now we can calculate the equilibrium concentrations of all substances: \[ [CO_2]_{eq} = [H_2]_{eq} = 0.50 - x = 0.50 - 0.217 = 0.283\, \text{M}\] \[ [CO]_{eq} = [H_2O]_{eq} = x = 0.217\, \text{M} \] The equilibrium concentrations are CO2: 0.283 M, H2: 0.283 M, CO: 0.217 M, and H2O: 0.217 M.

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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, denoted as \(K_c\), is a vital concept in understanding chemical equilibrium. In simple terms, it provides a mathematical relationship that allows us to predict the concentrations of reactants and products at equilibrium. For the reaction \(\text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g)\), the equilibrium constant expression is given by the formula:
\[\begin{aligned}K_c = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{CO}_2][\text{H}_2]}\end{aligned}\]
This ratio compares the concentrations of the products to the reactants, each raised to the power of their coefficients in the balanced chemical equation.
  • A large \(K_c\) indicates a reaction with more products than reactants at equilibrium.
  • A small \(K_c\) suggests few products compared to reactants.
Understanding \(K_c\) helps chemists to determine how far a reaction will proceed and to calculate the concentrations of substances at equilibrium.
ICE Table
An ICE table is a straightforward yet powerful tool used to organize and calculate the changes in concentration of species as a chemical reaction moves to equilibrium. The acronym "ICE" stands for Initial, Change, and Equilibrium:
  • Initial: The concentrations of all reactants and products before the reaction starts.
  • Change: The shift in concentration, usually represented by \(x\), as the reaction progresses.
  • Equilibrium: The concentrations after the reaction has reached equilibrium.
For example, in the reaction \(\text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g)\), the ICE table helps us keep track of concentrations:

- Initial: \([\text{CO}_2]_0 = 0.50 \, \text{M}\), \([\text{H}_2]_0 = 0.50 \, \text{M}\), while initially, \([\text{CO}]_0 = 0 \, \text{M}\), \([\text{H}_2\text{O}]_0 = 0 \, \text{M}\).
- Change: \(-x\) for \(\text{CO}_2\) and \(\text{H}_2\), \(+x\) for \(\text{CO}\) and \(\text{H}_2\text{O}\).
- Equilibrium: \(0.50-x\) for \(\text{CO}_2\) and \(\text{H}_2\), \(x\) for \(\text{CO}\) and \(\text{H}_2\text{O}\).The ICE table is crucial for setting up the equations needed to find \(x\), which in turn gives us the equilibrium concentrations.
Reaction Quotient
The reaction quotient, denoted as \(Q\), is an equation resembling the equilibrium constant expression but uses the initial concentrations instead of the equilibrium concentrations. It is a snapshot of the reaction at any stage, not just at equilibrium. Considering the reaction \(\text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g)\), the reaction quotient \(Q\) is formulated as:
\[\begin{aligned}Q = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{CO}_2][\text{H}_2]}\end{aligned}\]
By comparing \(Q\) to \(K_c\):
  • If \(Q = K_c\), the system is at equilibrium.
  • If \(Q < K_c\), the forward reaction is favored, going towards more products.
  • If \(Q > K_c\), the reverse reaction is favored, forming more reactants.
Thus, \(Q\) helps predict the direction in which the reaction will proceed to reach equilibrium. Understanding \(Q\) further aids in anticipating the behavior of a reaction under different initial conditions.

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Most popular questions from this chapter

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15 ), \(K_{p}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ is about \(1 \times 10^{-11}\). Assuming that the exhaust gas (total pressure \(101.3 \mathrm{kPa}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2},\) and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).

Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction \(4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\) in a fetus, compared to \(K_{c}\) for the same reaction in an adult.

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}(g)\) (b) \(\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SiCl}_{4}(g)\) (c) \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)\) (d) \(\mathrm{O}_{2}(g)+2 \mathrm{CO}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)\) (e) \(\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}^{+}(a q)\) (f) \(\mathrm{Fe}^{2+}(a q)+\mathrm{Ce}^{4+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)\) (g) \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

Consider the hypothetical reaction $$\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)+2 \mathrm{C}(g)$$ A flask is charged with \(100 \mathrm{kPa}\) of pure \(\mathrm{A}\), after which it is allowed to reach equilibrium at \(25^{\circ} \mathrm{C}\). At equilibrium, the partial pressure of \(\mathrm{B}\) is \(25 \mathrm{kPa}\). (a) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p} ?(\mathbf{c})\) What could we do to maximize the yield of \(\mathrm{B}\) ?

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