Complete Question
Sodium carbonate, Na2CO3(s), can be prepared by heating sodium bicarbonate, NaHCO3(s).
2NaHCO3(s) ------> Na2CO3(s) + CO2(g) + H2O(g) Kp = 0.23 at 100oC
If a sample of NaHCO3 is placed in an evacuated flask and allowed to achieve equilibrium at 100oC, what will the total gas pressure be?
Answer:
The value is [tex]P_T = 0.959 \ atm [/tex]
Explanation:
From the question we are told that
The equation is
[tex]2NaHCO_3_{(s)} \rightarrow Na_2CO_3_{(s)} + CO2_{(g)} + H2O_{(g)}[/tex]
Generally the equilibrium partial pressure is mathematically represented as
[tex]K_p = P_{CO_2} + P_{H_2O}[/tex]
Here
[tex] P_{CO_2} \ and \ P_{H_2O}[/tex] are partial pressures of [tex] CO_2 \ and \ H_2O [/tex] gases
Note: when writing the formula for equilibrium partial pressure we consider only the gas in the reaction
Let assume that
[tex]P_{CO_2} = a[/tex]
Generally [tex]P_{CO_2} = P_{H_2O}[/tex] since they are both products of the reaction
So
[tex]K_p = a^2 [/tex]
From the question we are told that [tex]K_p = 0.23 [/tex]
So
[tex] 0.23= a^2 [/tex]
=> [tex] a = \sqrt{0.23} [/tex]
=> [tex] a = \sqrt{0.23} [/tex]
=> [tex] a = 0.4796 \ atm [/tex]
So
[tex]P_{CO_2} = 0.4796 \ atm[/tex]
and
[tex]P_{H_2O} = 0.4796 \ atm [/tex]
So the total pressure is
[tex]P_T = 0.4796 + 0.4796[/tex]
=> [tex]P_T = 0.959 \ atm [/tex]
