Question
Write the equilibrium expression, and calculate Kp for the reaction under the conditions described:
2NaHCO3 (s) --> <--- Na2CO3 (s) + H2O (g) + CO2 (g)
100 g of solid NaHCO3 was placed in a previously evacuated rigid 5.00 L container and heated to 160 degrees C. Some of the original solid remained and the total pressure in the container was 7.76 atm when equilibrium was reached.
The answer is 15.1, I need to find how to get it, and how to write the expression.
2NaHCO3 (s) --> <--- Na2CO3 (s) + H2O (g) + CO2 (g)
100 g of solid NaHCO3 was placed in a previously evacuated rigid 5.00 L container and heated to 160 degrees C. Some of the original solid remained and the total pressure in the container was 7.76 atm when equilibrium was reached.
The answer is 15.1, I need to find how to get it, and how to write the expression.
Answers
The expression is
K<sub>p</sub> = p<sub>H2O</sub>*p<sub>CO2</sub>.
(The solids don't enter into a K<sub>p</sub>.)
The total pressure is 7.76 atm; therefore, the partial pressure of each gas is 1/2 that. Plug in 3.88 for each of the partial pressures and you have K<sub>p</sub>. I worked it out and came up with 15.0544 which I would round to 15.0 (I always round to the even number when the last digit is a 5.) Follow the advice of your prof on rounding.
K<sub>p</sub> = p<sub>H2O</sub>*p<sub>CO2</sub>.
(The solids don't enter into a K<sub>p</sub>.)
The total pressure is 7.76 atm; therefore, the partial pressure of each gas is 1/2 that. Plug in 3.88 for each of the partial pressures and you have K<sub>p</sub>. I worked it out and came up with 15.0544 which I would round to 15.0 (I always round to the even number when the last digit is a 5.) Follow the advice of your prof on rounding.