Strontium metal is responsible for the red color in fireworks. Fireworks manufacturers use strontium carbonate, which can be produced by combining strontium metal, graphite (C), and oxygen gas. The formation of one mole of SrCO3 releases 1.220 x 103 kJ of energy.
a) Write a balanced thermochemical equation for the reaction.
b) What is ƒ´H when 10.00 L of oxygen at 25 oC and 1.00 atm is used by the reaction?
Sr + C + 3/2O2 ==> SrCO3 + 1.22E3 kJ
Use PV = nRT and calculate n for O2 at the conditions listed, then convert to mols O2. g = mols x molar mass.
(1.22E3 kJ/48g O2) x g O2 used = ?
a) The balanced thermochemical equation for the reaction can be written as:
2Sr(s) + 2C(s) + 3O2(g) -> 2SrCO3(s)
b) To calculate ΔH (enthalpy change) when 10.00 L of oxygen at 25°C and 1.00 atm is used by the reaction, we need to use the ideal gas law and convert the given conditions to the conditions specified in the thermochemical equation.
First, let's convert the volume to moles of oxygen gas. We will use the ideal gas law formula:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Rearranging the formula to solve for n:
n = PV / RT
Substituting the given values:
P = 1.00 atm
V = 10.00 L
R = 0.0821 L·atm/(mol·K)
T = 25 + 273 = 298 K
n = (1.00 atm * 10.00 L) / (0.0821 L·atm/(mol·K) * 298 K)
n = 0.408 mol
From the balanced equation, we can see that 3 moles of oxygen gas are required for the reaction. Therefore, 0.408 mol of oxygen is 0.408/3 = 0.136 moles of the reaction.
The given enthalpy change is given per mole of strontium carbonate formed, and it's given as 1.220 x 10^3 kJ per mole.
ΔH = (1.220 x 10^3 kJ/mol) * (0.136 mol)
ΔH = 166.32 kJ
a) To write a balanced thermochemical equation for the reaction, we need to show the reactants and products with their respective coefficients.
Reactants:
Strontium metal (Sr)
Graphite (C)
Oxygen gas (O2)
Products:
Strontium carbonate (SrCO3)
The balanced thermochemical equation is:
2Sr + 2C + 3O2 ⟶ 2SrCO3
b) To calculate ∆H when 10.00 L of oxygen at 25 oC and 1.00 atm is used by the reaction, we can use the ideal gas law and the stoichiometry of the balanced equation.
First, let's convert the given volume of oxygen gas to moles using the ideal gas law equation:
PV = nRT
Where:
P = pressure (1.00 atm)
V = volume (10.00 L)
n = moles (to be determined)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (25 oC + 273.15 = 298.15 K)
Plugging in the values:
(1.00 atm) * (10.00 L) = n * (0.0821 L·atm/(mol·K)) * (298.15 K)
Solving for n:
n = (1.00 atm * 10.00 L) / (0.0821 L·atm/(mol·K) * 298.15 K)
n ≈ 4.07 moles of O2
From the balanced equation, we know that the formation of 2 moles of SrCO3 releases 1.220 x 10^3 kJ of energy.
So, for the formation of 1 mole of SrCO3, the energy released would be:
(1.220 x 10^3 kJ) / 2 ≈ 610 kJ
Therefore, for the formation of 4.07 moles of SrCO3 (as determined from the moles of O2):
∆H = 4.07 moles * 610 kJ/mole
∆H ≈ 2,483.7 kJ
Therefore, ∆H is approximately 2,483.7 kJ.