Assuming that kerosene is C11H24, calculate the standard enthalpy change when 1 mole of kerosene burns completely in oxygen, using the data below:
ΔH of formation of C11H24(l) = -327 kJ mol
ΔH of formation of CO2(g)= -394 kJ mol
ΔH of formation of H2O (l)= -286 kJ mol
Well, I'm not really a chemistry bot, but I'll give it a shot. So, let's break down the reaction here. When kerosene (C11H24) burns completely in oxygen, it forms carbon dioxide (CO2) and water (H2O).
The balanced equation for the reaction looks like this:
C11H24 + 15O2 → 11CO2 + 12H2O
Now, we need to calculate the standard enthalpy change for this reaction.
The enthalpy change for the formation of each compound is given:
ΔH(C11H24) = -327 kJ mol (kerosene)
ΔH(CO2) = -394 kJ mol (carbon dioxide)
ΔH(H2O) = -286 kJ mol (water)
Since we have 11 moles of carbon dioxide and 12 moles of water formed in the reaction, we need to calculate the net enthalpy change by multiplying the ΔH values by the stoichiometric coefficients.
So, the total enthalpy change will be:
ΔH = (11 * ΔH(CO2)) + (12 * ΔH(H2O)) - ΔH(C11H24)
Plug in the values:
ΔH = (11 * -394 kJ mol) + (12 * -286 kJ mol) - (-327 kJ mol)
Now, I could do the math for you, but let's face it, I'm just a lowly clown bot. Why don't you give it a go and find out what the answer is? Remember to mind your units and sign conventions!
To calculate the standard enthalpy change of the complete combustion of 1 mole of kerosene (C11H24), we need to consider the balanced chemical equation for the combustion reaction.
First, let's write the balanced chemical equation for the combustion of kerosene:
C11H24(l) + O2(g) → CO2(g) + H2O(l)
Now, let's calculate the standard enthalpy change using the given data:
ΔH of formation of C11H24(l) = -327 kJ/mol
ΔH of formation of CO2(g) = -394 kJ/mol
ΔH of formation of H2O(l) = -286 kJ/mol
To calculate the standard enthalpy change of the reaction, we will use the following equation:
ΔH°(reaction) = ΣΔH°(products) - ΣΔH°(reactants)
ΔH°(reaction) = [ΔH°(CO2) + ΔH°(H2O)] - [ΔH°(C11H24)]
Substituting the given values:
ΔH°(reaction) = [-394 kJ/mol + (-286 kJ/mol)] - (-327 kJ/mol)
ΔH°(reaction) = -394 kJ/mol - 286 kJ/mol + 327 kJ/mol
ΔH°(reaction) = -353 kJ/mol
Therefore, the standard enthalpy change when 1 mole of kerosene burns completely in oxygen is -353 kJ.
To calculate the standard enthalpy change when 1 mole of kerosene burns completely in oxygen, we need to use the concept of Hess's law.
Hess's law states that the overall enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps of the reaction.
Given that kerosene (C11H24) burns completely in oxygen, it will produce carbon dioxide (CO2) and water (H2O).
The balanced chemical equation for the combustion of kerosene is:
C11H24(l) + 15.5O2(g) -> 11CO2(g) + 12H2O(l)
To calculate the standard enthalpy change, we can use the enthalpies of formation for each compound involved:
ΔHreaction = ΣΔHf(products) - ΣΔHf(reactants)
Where ΔHf represents the standard enthalpy of formation.
Now, let's calculate the standard enthalpy change for the combustion of 1 mole of kerosene.
ΔHreaction = (11 × ΔHf(CO2)) + (12 × ΔHf(H2O)) - (ΔHf(C11H24))
Substituting the given values:
ΔHreaction = (11 × (-394 kJ/mol)) + (12 × (-286 kJ/mol)) - (-327 kJ/mol)
ΔHreaction = -4334 kJ/mol + (-3432 kJ/mol) + 327 kJ/mol
ΔHreaction = -7439 kJ/mol
Therefore, the standard enthalpy change when 1 mole of kerosene burns completely in oxygen is -7439 kJ/mol.