Calculate the heat of combustion of candle wax in kJ/mole. The standard heat of formation (∆Hf ) of candle wax (C20H42) is -2230 kJ/mole. The standard heat of formation (∆Hf ) of carbon dioxide is -394 kJ/mole. The standard heat of formation (∆Hf ) of water is -242 kJ/mole. The combustion of candle wax equals the ∆HRXN = (∆Hf (products) -(∆Hf (reactamts) ). Don’t forget to multiply the ∆Hf by the number of moles from the equation. Remember O2 ∆Hf = 0 because it is an element. Find the ∆HRXN.
Ah, the fascinating world of candle wax combustion! So, let's calculate the ∆HRXN (change in enthalpy) of the combustion of candle wax.
The equation for the combustion of candle wax (C20H42) is:
C20H42 + O2 → CO2 + H2O
Now, we need to calculate the reactants' ∆Hf (heat of formation). The wax is given as -2230 kJ/mol, but since we have 20 moles of C in the reactants, we'll multiply it by 20.
∆Hf (reactants) = 20 * -2230 kJ/mol = -44600 kJ/mol
The ∆Hf of O2 is 0 kJ/mol because it is an element.
Now, let's calculate the products' ∆Hf. We have CO2 and H2O, so we'll multiply their respective ∆Hf's by the number of moles from the equation.
∆Hf (CO2) = -394 kJ/mol * 20 moles = -7880 kJ/mol
∆Hf (H2O) = -242 kJ/mol * 21 moles = -5082 kJ/mol
Now, we can calculate the ∆HRXN:
∆HRXN = (∆Hf (products) - ∆Hf (reactants))
= [(-7880 kJ/mol) + (-5082 kJ/mol)] - (-44600 kJ/mol)
= -12962 kJ/mol - (-44600 kJ/mol)
= -12962 kJ/mol + 44600 kJ/mol
= 31638 kJ/mol
So, the ∆HRXN (heat of combustion) of candle wax is 31638 kJ/mol. But remember, this calculation was a bit dry. Need a candle pun to lighten the mood? Why did the candle go to college? Because it wanted to get a glowing education!
To calculate the ∆HRXN (enthalpy change of the reaction), we need to subtract the sum of the heat of formation (∆Hf) of the reactants from the sum of the heat of formation (∆Hf) of the products.
The balanced chemical equation for the combustion of candle wax (C20H42) can be written as:
C20H42 + 26O2 -> 20CO2 + 21H2O
Let's calculate the ∆HRXN step-by-step:
1. Calculate the ∆Hf (reactants):
- Candle wax (C20H42) has a ∆Hf of -2230 kJ/mole.
- Oxygen gas (O2) has a ∆Hf of 0 kJ/mole (since it is an element).
Therefore, the total ∆Hf of the reactants is (-2230 kJ/mole) + (0 kJ/mole) = -2230 kJ/mole.
2. Calculate the ∆Hf (products):
- Carbon dioxide (CO2) has a ∆Hf of -394 kJ/mole.
- Water (H2O) has a ∆Hf of -242 kJ/mole.
To determine the ∆Hf of the products, we need to multiply the ∆Hf values by the stoichiometric coefficients from the balanced equation:
20CO2: (20 mol) * (-394 kJ/mole) = -7880 kJ
21H2O: (21 mol) * (-242 kJ/mole) = -5082 kJ
Therefore, the total ∆Hf of the products is (-7880 kJ) + (-5082 kJ) = -12962 kJ.
3. Calculate the ∆HRXN:
∆HRXN = ∆Hf (products) - ∆Hf (reactants)
∆HRXN = -12962 kJ - (-2230 kJ)
∆HRXN = -10732 kJ/mole
Therefore, the heat of combustion of candle wax in kJ/mole (∆HRXN) is approximately -10732 kJ/mole.
To calculate the heat of combustion (∆HRXN) of candle wax (C20H42) in kJ/mole, we need to use the concept of standard heat of formation (∆Hf) of the reactants and products involved in the combustion reaction.
The balanced chemical equation for the combustion of candle wax can be written as:
C20H42 + 26O2 → 20CO2 + 21H2O
Now let's calculate ∆HRXN using the given standard heat of formation values:
Step 1: Calculate ∆Hf of the reactants:
∆Hf of C20H42 = -2230 kJ/mole (given)
∆Hf of O2 = 0 kJ/mole (since it is an element)
Step 2: Calculate ∆Hf of the products:
∆Hf of CO2 = -394 kJ/mole (given)
∆Hf of H2O = -242 kJ/mole (given)
Step 3: Multiply the ∆Hf values by the number of moles from the balanced equation:
-2230 kJ/mole × 1 mole (C20H42) = -2230 kJ
0 kJ/mole × 26 moles (O2) = 0 kJ
-394 kJ/mole × 20 moles (CO2) = -7880 kJ
-242 kJ/mole × 21 moles (H2O) = -5082 kJ
Step 4: Calculate ∆HRXN using the equation:
∆HRXN = (∆Hf of products) - (∆Hf of reactants)
∆HRXN = (-7880 kJ + (-5082 kJ)) - (-2230 kJ + 0 kJ)
∆HRXN = -12962 kJ + 2230 kJ
∆HRXN = -10732 kJ/mole
Therefore, the heat of combustion (∆HRXN) of candle wax is -10732 kJ/mole.