A 79.9-L sample of a gaseous hydrocarbon, measured at 1.00 atm pressure and 25.0°C,
is burned in excess oxygen, liberating 4.24 × 10^3 kJ of heat at constant pressure. What
is the identity of the hydrocarbon? (R = 0.0821 L · atm/(K · mol))
Substance H°f (kJ/mol)
CO2(g) –393.5
H2O(l) –285.8
A) ethane (C2H6, H°f = –84.68 kJ/mol)
B)B) ethylene (C2H4, H°f = 52.47 kJ/mol)
C) propane (C3H8, H°f = –104.7 kJ/mol)
D) acetylene (C2H2, H°f = 226.73 kJ/mol)
E) propylene (C3H6, H°f = 20.41 kJ/mol)
I would be happy to help you determine the identity of the hydrocarbon! Let me tell you a joke to lighten the mood first:
Why did the chemist lose their job?
Because they couldn't find the right solution!
Now onto the question. To find the identity of the hydrocarbon, we need to determine the heat of reaction (ΔH) for the combustion of the hydrocarbon using the heat of formation (ΔHf) values for the products.
The balanced equation for the combustion of a hydrocarbon is:
CnHm + (n + m/2)O2 → nCO2 + m/2H2O
We first need to calculate the moles of CO2 and H2O produced. Given that the volume is 79.9 L and the pressure is 1.00 atm, we can use the ideal gas law to find the number of moles of CO2:
PV = nRT
n = PV/RT
Since the pressure, volume, and temperature are given, we can substitute these values into the equation along with the value of the gas constant R, which is 0.0821 L·atm/(K·mol).
We can repeat the same calculation for H2O.
Next, we need to calculate the heat released by the combustion reaction. This can be done using the formula:
ΔH = Σ(ΔHf products) - Σ(ΔHf reactants)
We can substitute the values of ΔHf for CO2 and H2O into this equation. Given that the heat released is 4.24 × 10^3 kJ, we can calculate the value of ΔH for the reactant hydrocarbon.
Finally, we can compare the calculated ΔH of the hydrocarbon to the given values of ΔHf for the options provided to identify the correct hydrocarbon.
Remember, chemistry can be explosive, but it always has its solutions!
I hope this explanation helps you find the answer. Good luck!
To identify the hydrocarbon, we can use the heat of combustion equation and the given information. The heat of combustion (ΔHc) can be calculated using the equation:
ΔHc = n * ΔH(products) - ΔH(reactants)
where n is the stoichiometric coefficient of the hydrocarbon.
Let's calculate the moles of the hydrocarbon present in the 79.9 L sample using the Ideal Gas Law:
PV = nRT
n = PV / RT
n = (1.00 atm * 79.9 L) / (0.0821 L · atm/(K · mol) * (25.0 + 273) K)
n = 3.22 mol
Now, let's calculate ΔHc using the heat of formation values for the products and reactants:
ΔHc = n * (ΔH(CO2) + ΔH(H2O)) - 4.24 × 10^3 kJ
ΔHc = 3.22 mol * (-393.5 kJ/mol + -285.8 kJ/mol) - 4.24 × 10^3 kJ
ΔHc = -1250.94 kJ - 4.24 × 10^3 kJ
ΔHp = -5493.94 kJ
Now, let's check which hydrocarbon matches this value:
A) ethane: ΔHc = -84.68 kJ/mol * 3.22 mol = -272.29 kJ
B) ethylene: ΔHc = 52.47 kJ/mol * 3.22 mol = 168.89 kJ
C) propane: ΔHc = -104.7 kJ/mol * 3.22 mol = -337.53 kJ
D) acetylene: ΔHc = 226.73 kJ/mol * 3.22 mol = 730.27 kJ
E) propylene: ΔHc = 20.41 kJ/mol * 3.22 mol = 65.78 kJ
None of the options match the calculated ΔHc value of -5493.94 kJ. Therefore, none of the given hydrocarbons is the correct identity for the hydrocarbon.
To determine the identity of the hydrocarbon, we need to calculate the heat released from the combustion reaction and compare it to the enthalpies of formation of the potential hydrocarbons.
First, let's calculate the heat released from the combustion reaction using the given information:
q = -ΔH (since the heat is released)
q = -4.24 × 10^3 kJ
Now, let's calculate the number of moles of the hydrocarbon. We can use the ideal gas law equation, PV = nRT, to find the number of moles:
PV = nRT
n = PV / RT
Given:
P = 1.00 atm
V = 79.9 L
R = 0.0821 L · atm/(K · mol)
T = 25.0°C + 273.15 = 298.15 K
n = (1.00 atm) * (79.9 L) / (0.0821 L · atm/(K · mol) * 298.15 K)
Next, let's calculate the moles of CO2 and H2O produced in the combustion reaction. From the balanced equation, we know that one mole of hydrocarbon will produce one mole of CO2 and one mole of H2O.
Now, let's calculate the heat released per mole of hydrocarbon:
ΔH_per_mole = q / n
Finally, let's compare the calculated ΔH_per_mole to the enthalpies of formation (H°f) of the potential hydrocarbons and identify the hydrocarbon with the matching value.
The hydrocarbon that has an enthalpy of formation (H°f) value matching the calculated ΔH_per_mole is the identity of the hydrocarbon.
Let's calculate the answer using the formula and the given data.