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Chapter 13: Problem 116

At \(125^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ A \(1.00-\mathrm{L}\) flask containing 10.0 \(\mathrm{g} \mathrm{NaHCO}_{3}\) is evacuated and heated to \(125^{\circ} \mathrm{C} .\) a. Calculate the partial pressures of \(\mathrm{CO}_{2}\) and $\mathrm{H}_{2} \mathrm{O}$ after equilibrium is established. b. Calculate the masses of \(\mathrm{NaHCO}_{3}\) and $\mathrm{Na}_{2} \mathrm{CO}_{3}$ present at equilibrium. c. Calculate the minimum container volume necessary for all of the \(\mathrm{NaHCO}_{3}\) to decompose.

Short Answer

Expert verified
In summary, after the equilibrium is established at \(125^{\circ}\mathrm{C}\), the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) are 0.0297 atm. At equilibrium, there are 5.01 g of \(\mathrm{NaHCO}_{3}\) and 3.99 g of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) present. The minimum container volume required for all of the \(\mathrm{NaHCO}_{3}\) to decompose is 0.0141 L.

Step by step solution

01

Calculate the moles of NaHCO3

First, we need to find the moles of NaHCO3. We are given that the sample contains 10.0 g of NaHCO3. Using the molar mass of NaHCO3, we can find the moles: Moles of NaHCO3 = Mass of NaHCO3 / Molar mass of NaHCO3 The molar mass of NaHCO3 = 23 (Na) + 1 (H) + 12 (C) + 16 × 3 (O_3) = 84 g/mol Moles of NaHCO3 = 10.0 g / 84 g/mol = 0.119 moles
02

Set up the ICE table

Now that we have the moles of NaHCO3, we can set up an ICE (Initial, Change, Equilibrium) table to relate the amounts of all species involved. $$ \begin{array}{c|c|c|c|c} & \mathrm{Initial} & \mathrm{Change} & \mathrm{Equilibrium} \\ \hline \mathrm{NaHCO}_{3}(s) & 0.119 & -2x & 0.119-2x \\ \mathrm{Na}_{2}\mathrm{CO}_{3}(s) & 0 & x & x \\ \mathrm{CO}_{2}(g) & 0 & x & x \\ \mathrm{H}_{2}\mathrm{O}(g) & 0 & x & x \\ \end{array} $$
03

Use Kp to find the partial pressures

Now we can use the given Kp value (0.25) and the equilibrium amounts to find the partial pressures of CO2 and H2O. \(\mathrm{Kp}=\dfrac{[\mathrm{CO}_{2}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{NaHCO}_{3}]^2}=0.25\) Since the reaction is in a 1.00-L flask, the partial pressures will be equal to the molar concentrations. Thus, we can rewrite the equation as follows: \(0.25=\dfrac{x^2}{(0.119-2x)^2}\) To simplify the problem, we can assume that 2x is very small compared to 0.119. Thus, we can write the equation as: \(0.25=\dfrac{x^2}{(0.119)^2}\) Now, we can solve for x: \(x=\sqrt{0.25\times(0.119)^2}\) \(x=0.0297\) Now, we can find the partial pressures of CO2 and H2O: P(CO2) = P(H2O) = x = 0.0297 atm
04

Calculate the masses of NaHCO3 and Na2CO3 at equilibrium

Now that we have the amount of CO2 and H2O in moles at equilibrium, we can find the moles of NaHCO3 and Na2CO3 at equilibrium: Moles of NaHCO3 at equilibrium = 0.119 - 2x Moles of Na2CO3 at equilibrium = x Moles of NaHCO3 at equilibrium = 0.119 - 2(0.0297) = 0.0596 moles Moles of Na2CO3 at equilibrium = 0.0297 moles Now, we can find the masses of NaHCO3 and Na2CO3 at equilibrium: Mass of NaHCO3 = Moles of NaHCO3 × Molar mass of NaHCO3 Mass of Na2CO3 = Moles of Na2CO3 × Molar mass of Na2CO3 Mass of NaHCO3 = 0.0596 moles × 84 g/mol = 5.01 g Mass of Na2CO3 = 0.0297 moles × (46 + 12 + 48) g/mol = 3.99 g
05

Calculate the minimum container volume

To find the minimum container volume for all of the NaHCO3 to decompose, we can use the reaction stoichiometry and the Kp value. All of the NaHCO3 must decompose, so we will need to find the minimum volume that would allow the Kp value to be equal to 0.25. Let V be the required minimum volume. So, after the complete decomposition of 0.119 moles of NaHCO3: Moles of CO2 = Moles of H2O = Moles of Na2CO3 = 0.0595 moles Now, we can write the equation using Kp and V: \(0.25=\dfrac{[0.0595/V]^2}{[(\text{0.0595}/V)]^2}\) From here, we can solve for V: \(V=\dfrac{0.0595^2}{0.25}=0.0141\mathrm{L}\) So, the minimum container volume necessary for all of the NaHCO3 to decompose is 0.0141 L.

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