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Chapter 15: Problem 86

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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Most popular questions from this chapter

A mixture of \(1.374 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(70.31 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00-L vessel at \(700 \mathrm{~K}\). These substances react according to $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ At equilibrium, the vessel is found to contain \(0.566 \mathrm{~g}\) of \(\mathrm{H}_{2}\). (a) Calculate the equilibrium concentrations of $\mathrm{H}_{2}, \mathrm{Br}_{2},\( and \)\mathrm{HBr} .$ (b) Calculate \(K_{c}\)

Suppose that the gas-phase reactions \(A \longrightarrow B\) and $B \longrightarrow A\( are both elementary processes with rate constants of \)2.5 \times 10^{-2} \mathrm{~min}^{-1}\( and \)2.5 \times 10^{-1} \mathrm{~min}^{-1}$, respectively. (a) What is the value of the equilibrium constant for the equilibrium $\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?(\mathbf{b})$ Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

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) $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$

At \(1285^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is $K_{c}=1.04 \times 10^{-3} .\( A \)1.00-\mathrm{L}$ vessel containing an equilibrium mixture of the gases has \(1.50 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel?

At \(800 \mathrm{~K},\) the equilibrium constant for the reaction \(\mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{~A}(g)\) is $K_{c}=3.1 \times 10^{-4}$. (a) Assuming both forward and reverse reactions are elementary reactions, which rate constant do you expect to be larger, \(k_{f}\) or \(k_{r} ?\) (b) If the value of $k_{f}=0.27 \mathrm{~s}^{-1}\(, what is the value of \)k_{r}\( at \)800 \mathrm{~K} ?$ (c) Based on the nature of the reaction, do you expect the forward reaction to be endothermic or exothermic? (d) If the temperature is raised to $1000 \mathrm{~K}\(, will the reverse rate constant \)k_{r}$ increase or decrease? Will the change in \(k_{r}\) be larger or smaller than the change in \(k_{f}\) ?
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