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Question
Chemistry
Posted 7 months ago

8. ExtraCreds
At 125∘C,K4=0.25125^{\circ} \mathrm{C}, \mathrm{K}_{4}=0.25 for the reaction
2NaHCO3(s)⇔Na2CO2(s)+CO2(8)+H2O(8)
2 \mathrm{NaHCO}_{3}(s) \Leftrightarrow \mathrm{Na}_{2} \mathrm{CO}_{2}(s)+\mathrm{CO}_{2}(8)+\mathrm{H}_{2} \mathrm{O}(8)


A 1.00 L1.00 \mathrm{~L} flask containing 10.0 g10.0 \mathrm{~g} of NaHCO\mathrm{NaH} \mathrm{CO}, is evacusted and heated to 125∘C125^{\circ} \mathrm{C}
a. Calculate the partial pressure of CO2\mathrm{CO}_{2} And H2O\mathrm{H}_{2} \mathrm{O} after equalibrium is established
b. Calculate the masses of NaHCO\mathrm{NaHCO}, and Na2CO1\mathrm{Na}_{2} \mathrm{CO}_{1} present at equilibrium
c. Calculate the minimum container volume necessary for all the NaHCO2\mathrm{NaHCO}_{2} to decompose
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Answer from Sia
Posted 7 months ago
Solution
1
Convert mass of NaHCO₃ to moles: First, we need to convert the mass of sodium bicarbonate to moles using its molar mass. The molar mass of NaHCO₃ is 84.01 g/mol84.01 \, \text{g/mol}
$n = \frac{m}{M} = \frac{10.0 \, \text{g}}{84.01 \, \text{g/mol}} = 0.119 \, \text{moles}$
2
Write the equilibrium expression for Kp: The equilibrium constant in terms of partial pressures (Kp) is given by the equation Kp=PCO2⋅PH2O(PNaHCO3)2Kp = \frac{P_{CO_2} \cdot P_{H_2O}}{(P_{NaHCO_3})^2}. Since NaHCO₃ is a solid, its partial pressure is not included in the expression
3
Set up the ICE table and solve for partial pressures: Using an ICE (Initial, Change, Equilibrium) table, we can determine the changes in moles at equilibrium. Let x be the moles of NaHCO₃ that decompose
\begin{align*} \text{Initial:} & \quad 0.119 \, \text{moles NaHCO₃} & 0 \, \text{moles CO₂} & 0 \, \text{moles H₂O} \\ \text{Change:} & \quad -2x & +x & +x \\ \text{Equilibrium:} & \quad 0.119 - 2x & x & x \\ \end{align*}
4
Apply the Kp value to find x: Since the volume of the flask is 1.00 L, the partial pressures will be equal to the moles of gas at equilibrium. We can set up the equation for Kp using the equilibrium moles and solve for x
$0.25 = \frac{x \cdot x}{(0.119 - 2x)^2}$
5
Solve the quadratic equation for x: This will give us the moles of CO₂ and H₂O at equilibrium
After solving the quadratic equation, we find that $x \approx 0.042 \, \text{moles}$.
6
Calculate the partial pressures: The partial pressures of CO₂ and H₂O are equal to the moles of each at equilibrium since the volume is 1.00 L
$P_{CO_2} = P_{H_2O} = x = 0.042 \, \text{atm}$
1 Answer
The partial pressures of CO₂ and H₂O after equilibrium is established are both 0.042 atm.
Key Concept
The partial pressure of a gas in a mixture is directly proportional to its mole fraction in the mixture.
Explanation
At equilibrium, the partial pressures of CO₂ and H₂O are equal to the moles of each gas present in the 1.00 L flask.
Solution
1
Calculate the mass of NaHCO₃ remaining: Using the value of x, we can find the moles of NaHCO₃ that did not decompose
Moles of NaHCO₃ remaining = $0.119 - 2x = 0.119 - 2(0.042) = 0.035 \, \text{moles}$
2
Convert moles of NaHCO₃ to mass: Now we can convert the moles back to mass
Mass of NaHCO₃ = $0.035 \, \text{moles} \times 84.01 \, \text{g/mol} = 2.94 \, \text{g}$
3
Calculate the mass of Na₂CO₃ formed: For every mole of Na₂CO₃ formed, two moles of NaHCO₃ decompose
Mass of Na₂CO₃ = $x \times \text{molar mass of Na₂CO₃} = 0.042 \, \text{moles} \times 105.99 \, \text{g/mol} = 4.45 \, \text{g}$
2 Answer
The mass of NaHCO₃ remaining is 2.94 g, and the mass of Na₂CO₃ formed is 4.45 g.
Key Concept
The mass of a substance can be calculated from its moles by using its molar mass.
Explanation
The mass of NaHCO₃ and Na₂CO₃ at equilibrium can be determined by the moles of each substance and their respective molar masses.
Solution
1
Determine the moles of NaHCO₃ needed to decompose completely: To decompose all NaHCO₃, we need to find the total moles in 10.0 g
Total moles of NaHCO₃ = $0.119 \, \text{moles}$
2
Calculate the total moles of gas produced: When all NaHCO₃ decomposes, it will produce an equal amount of CO₂ and H₂O
Total moles of gas = $2 \times 0.119 \, \text{moles} = 0.238 \, \text{moles}$
3
Use the ideal gas law to find the minimum volume: At 125°C (398 K), we can use the ideal gas law PV=nRTPV = nRT to find the volume, where R is the gas constant 0.0821 \, \text{L·atm/(K·mol)}
$V = \frac{nRT}{P} = \frac{0.238 \, \text{moles} \times 0.0821 \, \text{L·atm/(K·mol)} \times 398 \, \text{K}}{1 \, \text{atm}} = 7.76 \, \text{L}$
3 Answer
The minimum container volume necessary for all the NaHCO₃ to decompose is 7.76 L.
Key Concept
The ideal gas law relates the pressure, volume, temperature, and number of moles of a gas.
Explanation
To find the minimum volume for complete decomposition, we use the ideal gas law with the total moles of gas that would be produced.

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