Question

At 125⁰C, K_p=0.25 for the reaction

$\mathbf{2}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{N}{\mathit{a}}_{\mathbf{2}}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

A 1.00-L flask containing 10.0 g of NaHCO3 is evacuated and heated to 125⁰C.

a. Calculate the partial pressures of CO₂ and H₂Oafter equilibrium is established.

b. Calculate the masses of NaHCO₃ and Na₂CO₃present at equilibrium.

c. Calculate the minimum container volume necessaryfor all the NaHCO₃ to decompose.

Short answer

(a) The partial pressures of CO₂ and H₂O after equilibrium is 0.5atm each.

(b) The masses of NaHCO₃ and Na₂CO₃ present at equilibrium are 2.5g and 1.6g respectively.

(c) The minimum container volume required will be 3.9L.

Step-by-step solution

Step 1:Determine partial pressures

(a)

$\begin{array}{l}2NaHC{O}_3\left(s\right)\rightleftharpoons N{a}_{2}C{O}_3\left(s\right)+C{O}_2\left(g\right)+H_{2}O\left(g\right)K_P=0.25\\ Initial--00\\ Change--+x+x\\ Equi--xx\end{array}$

Let the change be x.

$\begin{array}{c}{K}_P=0.25\\ K_P=P_{C{O}_2}\times P_{H_{2}O}\\ 0.25=P_{C{O}_2}\times P_{H_{2}O}\\ 0.25=x\times x\end{array}$

Given,

$\begin{array}{c}{x}^2=0.25\\ x=0.5\\ P_{C{O}_2}=0.5atm\\ P_{H_{2}O}=0.5atm\end{array}$

Determine the masses

(b)

Number of moles of CO₂ involved:

$\begin{array}{c}{n}_{C{O}_2}=\frac{P_{C{O}_2}V}{RT}\\ n_{C{O}_2}=\frac{\left(0.50atm\right)\left(1.00L\right)}{\left(0.08206Latm{K}^{-1}mo{l}^{-1}\right)\left(398K\right)}\\ n_{C{O}_2}=1.5\times {10}^{-2}molC{O}_2\end{array}$

Mass of$N{a}_{2}C{O}_3$ produced:

$\begin{array}{l}1.5\times {10}^{-2}molC{O}_2\times \frac{1molN{a}_{2}C{O}_3}{molC{O}_2}\times \frac{106.0gN{a}_{2}C{O}_3}{molN{a}_{2}C{O}_3}\\ =1.6gN{a}_{2}C{O}_3\end{array}$

Mass of$NaHC{O}_3$ reacted:

$\begin{array}{l}1.5\times {10}^{-2}molC{O}_2\times \frac{2molNaHC{O}_3}{molC{O}_2}\times \frac{84.01gNaHC{O}_3}{molNaHC{O}_3}\\ =2.5gNaHC{O}_3\end{array}$

Mass of $NaHC{O}_3$remaining:

$\begin{array}{l}=(10-2.5)g\\ =7.5g\end{array}$

Determineminimum container volume 

(c)

Number of moles of CO₂ involved

$\begin{array}{l}10.0gNaHC{O}_3\times \frac{1molNaHC{O}_3}{84.01gNaHC{O}_3}\times \frac{1molC{O}_2}{2molNaHC{O}_3}\\ =5.95\times {10}^{-2}molC{O}_2\end{array}$

When all NaHCO₃ is consumed we will havegas at a pressure of 0.50 atm.

Therefore, the volume required

$\begin{array}{c}V=\frac{nRT}{P}\\ =\frac{(5.95\times {10}^{-2}mol)\left(0.08206Latm{K}^{-1}mo{l}^{-1}\right)\left(398K\right)}{0.50atm}\\ =3.9L\end{array}$