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 species are: CO2: \( 0.184 \thinspace M \), H2: \( 0.184 \thinspace M \), CO: \( 0.316 \thinspace M \), and H2O: \( 0.316 \thinspace M \).

Step by step solution

01

Calculate initial concentrations

Calculate the initial concentrations of the reactants CO2 and H2, and assume that both products, CO and H2O, have an initial concentration of 0 mol/L since the reaction has not yet started. To do this, divide the given moles by the volume of the container (3.00 L). Initial concentrations: \[ [CO_2] = \frac{1.50 \thinspace mol}{3.00 \thinspace L} = 0.50 \thinspace M \] \[ [H_2] = \frac{1.50 \thinspace mol}{3.00 \thinspace L} = 0.50 \thinspace M \] \[ [CO] = [H_2O] = 0 \thinspace M \]
02

Set up the ICE table

Construct an ICE (Initial, Change, Equilibrium) table to show the changes in concentrations as the reaction reaches equilibrium. | | CO2 | + | H2 | ⇌ | CO | + | H2O | |---------------|-----|---|----|----|----|---|-----| | Initial (M) | 0.5 | | 0.5 | | 0 | | 0 | | Change (M) | -x | | -x | | +x | | +x | | Equilibrium |0.5-x| |0.5-x| | x | | x |
03

Write the equilibrium expression

Write the equilibrium expression (Kc) for the given reaction, and plug the equilibrium concentrations from the ICE table into the expression. \[ K_c = \frac{[CO][H_2O]}{[CO_2][H_2]} \] Given Kc = 0.802: \[ 0.802 = \frac{(x)(x)}{(0.5-x)(0.5-x)} \]
04

Solve for x

Solve the equation for x, which represents the change in concentration for each species as the reaction reaches equilibrium. Notice that this equation is a quadratic equation in the form of \( Ax^2 + Bx + C = 0 \). To solve the quadratic equation, you can use the quadratic formula: \[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \] However, in this case, it is reasonable to assume \( x \ll 0.5 \), so we can simplify the equation as follows: \[ 0.802 = \frac{x^2}{(0.5)(0.5)} \] Now solve for x: \[ x^2 = 0.802 \times (0.5)(0.5) \] \[ x = \sqrt{0.802 \times (0.5)(0.5)} \] \[ x \approx 0.316 \]
05

Calculate equilibrium concentrations

Now that we have the value of x, find the equilibrium concentrations of all the species by plugging x back into the values from the ICE table: \[ [CO_2]_{eq} = 0.5 - x = 0.5 - 0.316 \approx 0.184 \thinspace M \] \[ [H_2]_{eq} = 0.5 - x = 0.5 - 0.316 \approx 0.184 \thinspace M \] \[ [CO]_{eq} = x \approx 0.316 \thinspace M \] \[ [H_2O]_{eq} = x \approx 0.316 \thinspace M \] Equilibrium concentrations: CO2: 0.184 M H2: 0.184 M CO: 0.316 M H2O: 0.316 M

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Most popular questions from this chapter

Write the expression for \(K_{c}\) for the following reactions. \(\operatorname{In}\) each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) (b) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$ (c) \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) $\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)$ (e) $\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q)$ (f) $\mathrm{Fe}^{2+}(a q)+\mathrm{Zn}(s) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Fe}(s)$ (g) $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q)$

(a) Is the dissociation of fluorine molecules into atomic fluorine, \(\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{~F}(g)\) an exothermic or endothermic process? (b) If the temperature is raised by \(100 \mathrm{~K}\), does the equilibrium constant for this reaction increase or decrease? (c) If the temperature is raised by \(100 \mathrm{~K},\) does the forward rate constant \(k_{f}\) increase by a larger or smaller amount than the reverse rate constant \(k_{r} ?\)

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 5.00 -Lflaskat \(310 \mathrm{~K}\), \(40 \%\) of the $\mathrm{SO}_{2} \mathrm{Cl}_{2}\( decomposes to \)\mathrm{SO}_{2}\( and \)\mathrm{Cl}_{2}$ : $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{P}\) for this reaction at \(310 \mathrm{~K}\). (c) According to Le Châtelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture was transferred to a 1.00 -L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when 2.00 mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(1.00-\mathrm{L}\) vessel at \(310 \mathrm{~K}\).

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?

Ozone, \(\mathrm{O}_{3},\) decomposes to molecular oxygen in the stratosphere according to the reaction $2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)$. Would an increase in pressure favor the formation of ozone or of oxygen?

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