Sodium carbonate, Na2CO3(s), can be prepared by heating sodium bicarbonate, NaHCO3(s).
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) Kp = 0.34 at 160 ºC
If a sample of NaHCO3 is placed in an evacuated flask and allowed to achieve equilibrium at 100 ºC, what will the total gas pressure be?
Did you make typo? If Kp is at 160 C then it will be different at 100 C. Assuming you mean Ptotal at 160 C...
....2NaHCO3(s)=>Na2CO3(S)+CO2(g)+H2O(g)
I...solid.......solid......0......0
C...solid.......solid......p......p
E...solid.......solid......p......p
Kp = pCO2 * pH2O
Substitute the E line into Kp expression and solve for p. Then total P is 2p.
To find the total gas pressure at equilibrium, we need to determine the partial pressure of each gas in the equilibrium reaction. Since the equation provides the value of Kp, we can use it to find the partial pressures.
At equilibrium, the equation tells us that 2 moles of NaHCO3 will produce 1 mole of Na2CO3, 1 mole of CO2, and 1 mole of H2O.
Let's assume the initial moles of NaHCO3 is "x."
According to the equation, 2 moles of NaHCO3 would produce 1 mole of Na2CO3, CO2, and H2O. Therefore, at equilibrium, we would have x/2 moles of Na2CO3, CO2, and H2O.
Since the total moles of gases at equilibrium is (x/2) + (x/2) + (x/2) = 3x/2 moles.
Now, let's consider the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Using the ideal gas law, we can express the pressure of each gas in terms of partial pressure:
P(Na2CO3) = (1 mole)(R)(100°C + 273.15K) / V
P(CO2) = (1 mole)(R)(100°C + 273.15K) / V
P(H2O) = (1 mole)(R)(100°C + 273.15K) / V
Since the ideal gas constant "R" cancels out, we can rewrite the pressures in terms of moles and temperature only:
P(Na2CO3) = (1/2)x(100°C + 273.15K) / V
P(CO2) = (1/2)x(100°C + 273.15K) / V
P(H2O) = (1/2)x(100°C + 273.15K) / V
Since the total pressure is the sum of the partial pressures, we can find the total pressure (Ptotal) as follows:
Ptotal = P(Na2CO3) + P(CO2) + P(H2O)
Ptotal = [(1/2)x(100°C + 273.15K) / V] + [(1/2)x(100°C + 273.15K) / V] + [(1/2)x(100°C + 273.15K) / V]
We need to note that the temperature should be converted to Kelvin by adding 273.15 to it.
Therefore, the total gas pressure at equilibrium will be:
Ptotal = (3/2) [(100°C + 273.15K) / V]
Please note that we cannot determine the exact value of Ptotal without knowing the volume of the flask (V).
To find the total gas pressure at equilibrium, we need to calculate the partial pressure of each gas and then sum them up.
Given:
Reaction: 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
Kp = 0.34 at 160 ºC
We are asked to find the total gas pressure at equilibrium when the reaction occurs at 100 ºC. However, since the value of Kp is given at 160 ºC, we need to use the equation relating Kp and temperature to find the Kp value at 100 ºC.
The equation relating Kp and temperature is:
ln(Kp1/Kp2) = ΔHº/R * (1/T1 - 1/T2)
Where:
Kp1 and Kp2 are the equilibrium constants at temperatures T1 and T2 respectively.
ΔHº is the standard enthalpy change of the reaction.
R is the gas constant (8.314 J/(mol·K)).
We can rearrange this equation to solve for the Kp value at 100 ºC.
ln(Kp/0.34) = ΔHº/(8.314 J/(mol·K)) * (1/(273+160 K) - 1/(273+100 K))
ln(Kp/0.34) = ΔHº/(8.314 J/(mol·K)) * (0.0030603084 K⁻¹)
Kp/0.34 = e^(ΔHº/(8.314 J/(mol·K)) * (0.0030603084 K⁻¹))
Kp = 0.34 * e^(ΔHº/(8.314 J/(mol·K)) * (0.0030603084 K⁻¹))
Now, let's substitute the given values:
Kp = 0.34 * e^(ΔHº/(8.314 J/(mol·K)) * (0.0030603084 K⁻¹))
ΔHº = 0 since no temperature change is given
Kp = 0.34 * e^(0/(8.314 J/(mol·K)) * (0.0030603084 K⁻¹))
Kp = 0.34
Since Kp = 0.34 at 100 ºC, we can assume the reaction is close to equilibrium. This means that most of the NaHCO3 has been converted to Na2CO3, and the pressure of CO2 and H2O gases will be negligible.
Therefore, the total gas pressure will be close to zero since the only significant gas present is CO2, and its pressure is negligible.