I have the following marathon problem but am unsure where to start:
A Sample consisting of 22.7g of nongaseous, unstable compound X is placed inside a metal cylinder with a radius of 8.00cm, and a piston is carefully placed on the surface of the compound so that, for all practical 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 plugged so that nothing inside the cylinder can escape.) The piston-and-cylinder apparatus is carefully placed in 10.00 kg of water at 25.00oC. The barometric pressure is 778 torr.
When the compound spontaneously decomposes, the piston moves up, the temperature of the water reaches a maximum of 29.52oC, 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 carbon dioxide, 0.250 mol liquid water, 0.025 mol oxygen gas, and an undetermined amount of 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 4.184 J/oC · g. The conversion factor between L · atm and J can be determined from the two values for the gas constant R, namely, 0.08206 L · atm/mol · K and 8.3145 J/mol · K. The vapor pressure of water at 29.5 oC 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 the 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.
See my response above.
To solve this marathon problem, let's break it down step by step:
A. Determining the formula for X:
We are given that the cylinder contains 0.300 mol of carbon dioxide (CO2), 0.250 mol of liquid water (H2O), 0.025 mol of oxygen gas (O2), and an undetermined amount of gaseous element A. To find the formula for X, we need to consider the total number of moles of each element present.
From the given information, we have:
Carbon (C): 1 atom in 0.300 mol CO2
Hydrogen (H): 2 atoms in 0.250 mol H2O
Oxygen (O): 2 atoms in 0.250 mol H2O + 1 atom in 0.025 mol O2
Now, let's calculate the total moles of each element:
C: 0.300 mol
H: 0.500 mol
O: 0.300 mol + 0.025 mol = 0.325 mol
The formula for X can be determined by finding the simplest whole number ratio of moles of each element. Divide each element's moles by the smallest mole value (0.300):
C: 0.300 mol / 0.300 mol = 1
H: 0.500 mol / 0.300 mol ≈ 1.67, which we can approximate to 2
O: 0.325 mol / 0.300 mol ≈ 1.08, which we can approximate to 1
Therefore, the formula for X is COH2O.
B. Calculating the pressure-volume work for the decomposition of X:
To calculate the work done by the system, we can use the formula:
Work = -(pressure × change in volume)
We know that the pressure is given by the barometric pressure, which is 778 torr. We need to convert this pressure to atm:
1 atm = 760 torr
So, the pressure is approximately 778 torr / 760 = 1.024 atm.
The change in volume can be found by subtracting the initial volume from the final volume. The initial volume is the volume of the cylinder, which can be calculated using the formula for the volume of a cylinder:
Volume = π × radius^2 × height
Given:
Radius = 8.00 cm = 8.00 cm × (1 m / 100 cm) = 0.08 m
Height = 59.8 cm = 59.8 cm × (1 m / 100 cm) = 0.598 m
Volume = π × 0.08 m^2 × 0.598 m = 0.151 L
The final volume is essentially zero, as the piston is at the bottom of the cylinder. So the change in volume is -0.151 L.
Now, we can calculate the work:
Work = -(1.024 atm × -0.151 L) = 0.155424 L·atm or 155.42 J
To convert the work from joules to kilojoules, divide by 1000:
Work = 155.42 J / 1000 = 0.155 kJ
Therefore, the pressure-volume work for the decomposition of the 22.7 g sample of X is approximately 0.155 kJ.
C. Finding the molar change in internal energy and the approximate standard enthalpy of formation for X:
The molar change in internal energy can be calculated using the equation:
ΔU = q + w
where ΔU represents the molar change in internal energy, q is the heat transferred, and w is the work done by the system.
We have already found the work in part B. Now, let's calculate the heat transfer (q) using the equation:
q = m × C × ΔT
where m is the mass of water, C is the heat capacity of water, and ΔT is the change in temperature of water.
Given:
Mass of water (m) = 10.00 kg
Heat capacity of water (C) = 4.184 J/oC·g
Change in temperature (ΔT) = 29.52oC - 25.00oC = 4.52oC
q = (10.00 kg) × (4.184 J/oC·g) × (4.52oC) = 1872.9536 J
q ≈ 1873 J
Next, we can calculate the molar change in internal energy:
ΔU = q + w
ΔU = 1873 J + 155.42 J = 2028.42 J
To convert the molar change in internal energy from joules to kilojoules, divide by 1000:
ΔU = 2.02842 kJ
The standard enthalpy change (ΔH) for the decomposition of X is given as -1893 kJ/mol X.
By definition, ΔH = ΔU + PΔV, where P is the external pressure and ΔV is the change in volume. In this case, ΔV is essentially zero since the final volume is negligible, and the external pressure (P) is the same as the barometric pressure (778 torr). Therefore, we can approximate the standard enthalpy of formation (ΔHf) for X as:
ΔHf ≈ ΔH = -1893 kJ/mol
Thus, the approximate standard enthalpy of formation for X is -1893 kJ/mol.
To summarize:
A. The formula for X is COH2O.
B. The pressure-volume work for the decomposition of the 22.7 g sample of X is approximately 0.155 kJ.
C. The molar change in internal energy for the decomposition of X is approximately 2.02842 kJ/mol X, and the approximate standard enthalpy of formation for X is -1893 kJ/mol.