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?
Kp = 0.23 = pCO2*pH2O
pCO2 = pH2O
p = sqrt(0.23) = ?
pCO2 + pH2PO = ?
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?
A) 0.46 atm B) 0.96 atm C) 0.23 atm D) 0.48 atm E) 0.11 atm
Well, let me crunch some numbers for you! According to the equation, for every 2 moles of NaHCO3 that react, we get 1 mole of Na2CO3, 1 mole of CO2, and 1 mole of H2O.
Now, since we're dealing with an equilibrium, we have to consider the fact that some of the reactants will react and some products will reverse react. But fear not, the equilibrium constant, Kp, tells us the ratio of the partial pressures of the products and reactants at equilibrium.
In this case, the Kp value is 0.23. Since the ratio of the moles of products to reactants is the same as the ratio of their partial pressures, we can say that the partial pressure of the CO2 and H2O gases will be 0.23 times the partial pressure of the Na2CO3 solid.
But wait! We need to find the total gas pressure. So we need to find the partial pressure of CO2 and H2O and then add them together. Since they both have the same partial pressure, we can find it by taking the square root of the Kp value.
So, the partial pressure of CO2 (and H2O) is approximately √0.23. Now, to find the total gas pressure, we add this to the partial pressure of Na2CO3.
But oh, silly me! I forgot to ask you the amount of NaHCO3 that was placed in the flask. Without that information, I can't calculate the partial pressures accurately. So, please provide the amount, and I'll be happy to help you crunch the numbers and give you the total gas pressure.
To find the total gas pressure at equilibrium, we need to use the given equilibrium constant, Kp, and the balanced equation.
The balanced equation for the reaction is:
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
From the equation, we can see that for every two moles of NaHCO3, we get one mole of CO2 and one mole of H2O.
Considering the stoichiometry, we can say that when two moles of NaHCO3 react, they produce one mole of CO2 and one mole of H2O. So, the mole ratio of CO2 to NaHCO3 is 1:2.
Now, let's consider the equation for Kp:
Kp = (P_CO2 * P_H2O) / (P_Na2CO3 * P_NaHCO3^2)
Here, P_CO2 represents the partial pressure of CO2, P_H2O represents the partial pressure of H2O, P_Na2CO3 represents the partial pressure of Na2CO3, and P_NaHCO3 represents the partial pressure of NaHCO3.
Since the flask is evacuated, the initial partial pressures of CO2, H2O, Na2CO3, and NaHCO3 are all zero.
At equilibrium, we assume that the partial pressures of CO2, H2O, Na2CO3, and NaHCO3 are P_CO2, P_H2O, P_Na2CO3, and P_NaHCO3, respectively.
Since the reaction stoichiometry tells us that two moles of NaHCO3 produce one mole of CO2, we can write P_CO2 = (1/2) * P_NaHCO3.
Using the equilibrium expression for Kp, we can rewrite it as:
Kp = (P_CO2 * P_H2O) / (P_Na2CO3 * P_NaHCO3^2)
= ((1/2) * P_NaHCO3 * P_H2O) / (P_Na2CO3 * P_NaHCO3^2)
Since we have only one solid (Na2CO3) and we're assuming an evacuated flask, the partial pressure of Na2CO3, P_Na2CO3, is also considered to be zero.
Now, rearranging the equation, we get:
P_NaHCO3^2 = ((1/2) * P_NaHCO3 * P_H2O) / Kp
Simplifying further, we get:
P_NaHCO3 = (2 * Kp * P_Na2CO3 * P_NaHCO3^2) / P_H2O
Simplifying once more, we get:
P_NaHCO3^3 = (2 * Kp * P_NaHCO3) / P_H2O
Here, we can substitute the given value of Kp, which is 0.23, and assume the partial pressure of water (P_H2O) is 1 (since it's in the gas phase at 100ºC). So, our equation becomes:
P_NaHCO3^3 = (2 * 0.23 * P_NaHCO3) / 1
Now, solving for P_NaHCO3, we get:
P_NaHCO3^3 = 0.46 * P_NaHCO3
Dividing both sides by P_NaHCO3, we get:
P_NaHCO3^2 = 0.46
Taking the square root of both sides, we get:
P_NaHCO3 = √0.46
Finally, we can find the total gas pressure at equilibrium by using the ideal gas law, which states:
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.
In this case, we are given the temperature as 100ºC, which is the same as 373K. We also know that the number of moles of CO2 and H2O is equal because of the stoichiometry, so we can consider them together as one mole.
Using the given equation,
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
we can say that two moles of NaHCO3 produce one mole of CO2 and one mole of H2O.
Thus, the total number of moles of gas at equilibrium is 1 (CO2 + H2O).
Now, substituting the values into the ideal gas law equation, we get:
P * V = n * R * T
P * 1L = 1 * R * 373K
Solving for P, we find:
P = R * 373K
Using the ideal gas constant, R, which is approximately 0.0821 L.atm/(mol.K), we calculate:
P ≈ 0.0821 L.atm/(mol.K) * 373K
Thus, the total gas pressure at equilibrium is approximately 30.6 atm.