When 34.1 g of lead reacts with 6.81 L of oxygen gas, measured at 1.00 atm and 25.0°C, 36.1 kJ of heat is released at constant pressure. What is DH° for this reaction?
What reaction? The most plausible answer for a question that leaves me with questions is dH rxn = -36.1 kJ since it is exothermic.
Well, this reaction seems to be quite "hot"!
To calculate ΔH°, we need to use the formula:
ΔH° = q / n
Where q is the heat released and n is the number of moles of the limiting reactant.
First, let's calculate the number of moles of oxygen gas using the ideal gas law:
PV = nRT
n = PV / RT
Given that P = 1.00 atm, V = 6.81 L, R = 0.0821 L·atm/(mol·K), and T = 25.0°C (298 K), we can calculate n.
Now, we need to figure out the number of moles of lead. We can use its molar mass to do that.
Finally, we divide the heat released, 36.1 kJ, by the number of moles of the limiting reactant (either oxygen or lead) to get the answer.
Just remember, my jokes are truly exothermic!
To find the standard enthalpy change (ΔH°) for the reaction, we need to use the following equation:
ΔH° = q / n
Where:
ΔH° = Standard enthalpy change
q = Heat released or absorbed in the reaction
n = Number of moles of the limiting reactant
First, we need to determine the limiting reactant between lead (Pb) and oxygen (O2). To do this, we can calculate the number of moles for each reactant.
The molar mass of lead (Pb) is 207.2 g/mol. So, the number of moles of Pb is:
n(Pb) = mass / molar mass
n(Pb) = 34.1 g / 207.2 g/mol
n(Pb) = 0.1646 mol
The molar volume of a gas at STP is 22.4 L/mol. Since the given volume of oxygen gas is not at STP, we need to use the ideal gas law to calculate the number of moles of oxygen.
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in L)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature (in Kelvin)
First, convert the given temperature from Celsius to Kelvin:
T = 25.0°C + 273.15 = 298.15 K
Now, rearrange the equation to solve for n(O2):
n(O2) = PV / RT
n(O2) = (1.00 atm) * (6.81 L) / (0.0821 L·atm/mol·K * 298.15 K)
n(O2) ≈ 0.282 mol
The ratio of Pb to O2 in the balanced chemical equation is 4:1. Since the number of moles of oxygen (0.282 mol) is greater than the number of moles of lead (0.1646 mol), we can conclude that oxygen is the limiting reactant.
Therefore, we will use the number of moles of oxygen (0.1646 mol) to calculate ΔH°.
We are given that q = -36.1 kJ (negative because the reaction releases heat).
ΔH° = q / n(O2)
ΔH° = -36.1 kJ / 0.1646 mol
ΔH° ≈ -219.44 kJ/mol
So, the standard enthalpy change (ΔH°) for this reaction is approximately -219.44 kJ/mol.
To calculate the standard enthalpy change (ΔH°) for the given reaction, we can use the equation:
ΔH° = q / n
Where:
- ΔH° is the standard enthalpy change
- q is the heat transferred (in this case, -36.1 kJ since heat is released)
- n is the number of moles of the limiting reactant
First, let's determine the number of moles of the limiting reactant. We'll use the ideal gas law to calculate the number of moles of oxygen gas (O2):
PV = nRT
P = 1.00 atm (pressure)
V = 6.81 L (volume)
n = ?
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 25.0°C = 298 K (temperature in Kelvin)
To solve for n, rearrange the equation:
n = PV / RT
Substituting the values, we have:
n = (1.00 atm)(6.81 L) / (0.0821 L·atm/(mol·K))(298 K)
n ≈ 0.275 mol
Now that we know the number of moles of oxygen gas, we can calculate the ΔH°:
ΔH° = q / n
ΔH° = -36.1 kJ / 0.275 mol
ΔH° ≈ -131.6 kJ/mol
Therefore, the standard enthalpy change (ΔH°) for this reaction is approximately -131.6 kJ/mol.