A sample consisting of 22.7 g of a nongaseous, unstable compound X is placed inside a metal cylinder with a radius of 8.00 cm, and a piston is carefully placed on the surface of the compound so that, for all practicla purposes, the distance between the bottom of the cylinder and the piston is zero. (A hole in the piston allows trapped air to escape as the piston is placed on the compound; then this hole is pulugged so that nothing inside the cylinder can escape.) The piston-and-cylinder apparatus is carefully placed in 10.00kg of water at 25.00 degrees C. The barometric pressure is 778 torr.
When the compound spontaneously decomposes, the piston moves up, the temperature of the water reaches a maximum of 29.52 C, and then it gradually decreases as the water loses heat to the surrounding air. The distance between the piston and the bottom of the cylinder, at the maximum temperature, is 59.8 cm. Chemical analysis shows that the cylinder contains 0.300 mol carbond dioxide, 0.250 mol liquid water, 0.025 mol oxygen gas, and an undetermined amount of a gaseous element A.
It is known that the enthalpy change for the decomposition of X, according to the reaction described above, is -1893 KJ/mol X. The standard enthalpies of formation for gaseous carbon dioxide and liquid water are -393.5 KJ/mol and -286 KJ/mol, respectively. The heat capacity for water is 4.184J/*C (times) g. The conversion factor between L (times) atm and J can be determined from the two calues for the gas constant R, namely, 0.08206 L(times)atm/mol(times)K and 8.3145 J/mol (times) K. The vapor pressure of water at 29.5 *C is 31 torr. Assume that the heat capacity of the piston-and-cylinder apparatus is negligible and that the piston has negligible mass.
Given the preceding information, determine
a. The formula for X
b. The pressure-volume work (in KJ)for the decomposition of 22.7-g sample of X.
c. The molar change in internal energy for the decomposition of X and the approximate standard enthalpy of formation for X.
To determine the answers to the given questions, we need to follow a step-by-step approach and use the information provided. Let's break it down:
a. The formula for X:
The given information states that the compound X decomposes to form carbon dioxide (CO2), water (H2O), and oxygen (O2). From the mole ratios provided, we can calculate the moles of each component:
- Carbon dioxide (CO2): 0.300 mol
- Water (H2O): 0.250 mol
- Oxygen (O2): 0.025 mol
To determine the formula for X, we need to consider the difference in moles between the reactant (X) and the products (CO2, H2O, O2). Since carbon dioxide and water are formed in equal moles, X must have the same number of moles before decomposition. Therefore, the formula for X is C0.300H0.250.
b. The pressure-volume work for the decomposition of X:
To calculate the pressure-volume work, we need to use the equation:
Work = -PΔV,
where ΔV is the change in volume and P is the pressure.
We are given the following information:
- The distance between the piston and the bottom of the cylinder is 59.8 cm.
- The radius of the cylinder is 8.00 cm.
To determine ΔV, we need to calculate the initial and final volumes:
Initial volume (V1) = πr^2h1, where r is the radius and h1 is the initial distance between the piston and the bottom of the cylinder.
Final volume (V2) = πr^2h2, where r is the radius and h2 is the final distance between the piston and the bottom of the cylinder.
Using the given values:
- r = 8.00 cm = 0.08 m
- h1 = 0 (for all practical purposes)
- h2 = 59.8 cm = 0.598 m
Now, we can calculate V1 and V2:
V1 = π(0.08^2)(0) = 0
V2 = π(0.08^2)(0.598) = 0.01524 m^3
Next, we need to determine the pressure (P) at the final state. The barometric pressure provided is 778 torr. The pressure inside the piston-and-cylinder apparatus is the sum of the barometric pressure and the vapor pressure of water at 29.5°C:
Total Pressure (P) = 778 torr + 31 torr = 809 torr
Now, we can calculate the pressure-volume work:
Work = -PΔV = -(809 torr)(0.01524 m^3)
To convert torr to atm, we use the conversion factor: 1 atm = 760 torr.
Therefore, the work can be calculated as:
Work = -[(809 torr)/(760 torr/atm)](0.01524 m^3) = -0.01620 atm⋅m^3
Finally, since the work is expressed in Kilojoules (KJ), we need to convert it to KJ:
Work (KJ) = -0.01620 atm⋅m^3 × (101.325 J)/(1 atm⋅L) × (1 L)/(1000 m^3) × (1 KJ)/(1000 J)
c. The molar change in internal energy for the decomposition of X and the approximate standard enthalpy of formation for X:
The molar change in internal energy (ΔU) is related to the enthalpy change (ΔH) by the equation:
ΔU = ΔH - PΔV,
where PΔV is the work done.
Given:
- The enthalpy change for the decomposition of X (ΔH) = -1893 KJ/mol X
- The work done (PΔV) calculated in part b
To calculate ΔU, we can substitute the known values into the equation mentioned above:
ΔU = -1893 KJ/mol X - (work calculated in part b)
The approximate standard enthalpy of formation for X can be calculated by rearranging the equation:
ΔH = ΔU + PΔV
Substitute the known values:
ΔH = ΔU + (work calculated in part b)
Solve for ΔH to obtain the approximate standard enthalpy of formation for X.
Now you should have the necessary information to answer the questions (a, b, c) based on the given data and calculations explained above.