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Chapter 13: Problem 64

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ If 2.0 moles of \(\mathrm{CO}_{2}\) is initially placed into a 5.0 -L vessel, calculate the equilibrium concentrations of all species.

Short Answer

Expert verified
The equilibrium concentrations of the species are approximately: \([\text{CO}_{2}] \approx 0.397 \, \text{mol/L}\), \([\text{CO}] \approx 1.06 \times 10^{-3} \, \text{mol/L}\), and \([\text{O}_{2}] \approx 5.3 \times 10^{-4} \, \text{mol/L}\).

Step by step solution

01

Set up ICE table

First, we need to create an ICE (Initial, Change, Equilibrium) table with the given information. In this table, the initial concentrations of CO and O\(_2\) are not given, so we'll assume they are 0. \[ \begin{array}{|c|c|c|c|} \hline \textbf{Species} & \boldsymbol{[\textbf{CO}_{2}]} &\boldsymbol{[\textbf{CO}]} &\boldsymbol{[\textbf{O}_{2}]} \\ \hline \textbf{Initial} & \frac{2.0 \text{mol}}{5.0 \text{L}} & 0 & 0 \\ \hline \textbf{Change} & -2x & +2x & +x \\ \hline \textbf{Equilibrium} & \frac{2.0 \text{mol}}{5.0 \text{L}}-2x & 2x & x \\ \hline \end{array} \] Here, x represents the change in concentration of each species during the process of reaching equilibrium.
02

Write the equilibrium expression

Next, we need to write the equilibrium constant expression using the concentrations from the ICE table, and the given value of K: \(K = \frac{[\text{CO}]^{2}[\text{O}_{2}]}{[\text{CO}_{2}]^{2}}\)
03

Substitute equilibrium concentrations into the equilibrium expression

Now, replace the species' concentrations from the ICE table with the equilibrium concentrations in terms of x: \(2.0 \times 10^{-6} = \frac{(2x)^{2}(x)}{(\frac{2.0 \text{mol}}{5.0 \text{L}}-2x)^{2}}\)
04

Solve for x

At this point, we need to solve the equation for x. This might be a little bit complicated to solve directly; however, we can make some assumptions to simplify the calculation: As K is very small (2.0 x 10^{-6}), the reaction will not move significantly towards products. Thus, the values of 2x and x will be much smaller compared to 2.0 mol/5.0 L, which lets us ignore the 2x in the denominator: \(2.0 \times 10^{-6} \approx \frac{(2x)^{2}(x)}{(\frac{2.0 \text{mol}}{5.0 \text{L}})^{2}}\) Now, simplify and solve for x: \(x = \sqrt[3]{2.0 \times 10^{-6} \cdot (\frac{2.0 \text{mol}}{5.0 \text{L}})^{2}}\) \(x \approx 5.3 \times 10^{-4} \, \text{mol/L}\)
05

Calculate equilibrium concentrations

Finally, calculate the equilibrium concentrations of all species using the value of x: -\([CO_{2}] = \frac{2.0 \text{mol}}{5.0 \text{L}}-2x = \frac{2.0 \text{mol}}{5.0 \text{L}}-2(5.3 \times 10^{-4} \, \text{mol/L}) \approx 0.397 \, \text{mol/L}\) -\([CO] = 2x = 2(5.3 \times 10^{-4} \, \text{mol/L}) \approx 1.06 \times 10^{-3} \, \text{mol/L}\) -\([O_{2}] = x = 5.3 \times 10^{-4} \, \text{mol/L}\) So, the equilibrium concentrations are approximately: \( \begin{aligned} [\text{CO}_{2}] &= 0.397\, \text{mol/L}\\ [\text{CO}] &= 1.06 \times 10^{-3}\, \text{mol/L}\\ [\text{O}_{2}] &= 5.3 \times 10^{-4}\, \text{mol/L} \end{aligned} \)

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C}\) , equilibrium is reached when 50.0\(\%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

The following equilibrium pressures at a certain temperature were observed for the reaction $$\begin{aligned} 2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\ P_{\mathrm{NO}_{2}} &=0.55 \mathrm{atm} \\\ P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{atm} \\ P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{atm} \end{aligned}$$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

Suppose a reaction has the equilibrium constant \(K=1.7 \times 10^{-8}\) at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?

A 1.00-L flask was filled with 2.00 moles of gaseous \(\mathrm{SO}_{2}\) and 2.00 moles of gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that 1.30 moles of gaseous NO was present. Assume that the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ occurs under these conditions. Calculate the value of the equilibrium constant, \(K,\) for this reaction.

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is $$\mathrm{HCO}_{3}^{-}(a q) \leftrightharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) K=5.6 \times 10^{-11}$$ If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into 1.00 \(\mathrm{L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}^{2-2}\) ?
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