Calculate the change in enthalpy for the following reaction:
C25H52(g) + O2(g) → CO2(g) + H2O(l)
I started off by balancing the equation:
C25H52(g) + 38O2(g) → 25CO2(g) + 26H2O(l)
Can someone help me with the rest.
dHfrxn = (n*dHf products) - (n*dH reactants) = ?
Certainly! To calculate the change in enthalpy for a reaction, you can use the concept of Hess's Law, which states that the change in enthalpy for a reaction is equal to the sum of the change in enthalpy for each individual step of the reaction.
Here's how you can calculate the change in enthalpy for the given reaction:
1. First, you need to determine the change in enthalpy for each species in the reaction. This information is usually provided in thermochemical tables.
2. Write down the balanced chemical equation for each individual step of the reaction. In this case, the reaction is already balanced as:
C25H52(g) + 38O2(g) → 25CO2(g) + 26H2O(l)
3. Now, calculate the change in enthalpy (ΔH) for each step using the provided thermochemical data. The ΔH value represents the change in enthalpy for each mole of the substance involved in the reaction.
For example, let's say the ΔH for the combustion of C25H52(g) to form CO2(g) is -1400 kJ/mol and the ΔH for the formation of H2O(l) is -285.8 kJ/mol.
Using these values, you can calculate the total change in enthalpy for the reaction by multiplying the ΔH values by the stoichiometric coefficients and summing them up:
(25 mol CO2 × ΔH for CO2) + (26 mol H2O × ΔH for H2O) - (1 mol C25H52 × ΔH for C25H52) - (38 mol O2 × ΔH for O2)
4. Substitute the ΔH values:
(25 mol CO2 × ΔH for CO2) + (26 mol H2O × ΔH for H2O) - (1 mol C25H52 × ΔH for C25H52) - (38 mol O2 × ΔH for O2)
= (25 mol CO2 × ΔH for CO2) + (26 mol H2O × ΔH for H2O) - (1 mol C25H52 × (-1400 kJ/mol)) - (38 mol O2 × ΔH for O2)
5. Multiply stoichiometric coefficients with their respective ΔH values:
= (25 mol CO2 × ΔH for CO2) + (26 mol H2O × ΔH for H2O) - (-1400 kJ/mol × 1 mol C25H52) - (38 mol O2 × ΔH for O2)
= (25 mol CO2 × ΔH for CO2) + (26 mol H2O × ΔH for H2O) + (1400 kJ/mol × 1 mol C25H52) - (38 mol O2 × ΔH for O2)
6. Substitute the values for the ΔHs provided in the thermochemical tables to find the final result:
= (25 mol CO2 × ΔH for CO2) + (26 mol H2O × ΔH for H2O) + (1400 kJ/mol × 1 mol C25H52) - (38 mol O2 × ΔH for O2)
By substituting the actual numerical values for the ΔHs, you can calculate the overall change in enthalpy for the given reaction.
To calculate the change in enthalpy (ΔH) for the given reaction, you need to use the standard enthalpy of formation (ΔHf) values for the reactants and products. The ΔHf values represent the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states.
Here are the balanced equations for the formation of each compound:
C25H52(g) + 38O2(g) → 25CO2(g) + 26H2O(l)
Now, you will need to look up the ΔHf values for each compound in a reliable source. Here are the ΔHf values at 25°C (298 K):
ΔHf(C25H52(g)) = 0 kJ/mol
ΔHf(CO2(g)) = -393.5 kJ/mol
ΔHf(H2O(l)) = -285.8 kJ/mol
Next, you can use the following equation to calculate the change in enthalpy:
ΔH = Σ (ΔHf of products) - Σ (ΔHf of reactants)
Now plug in the ΔHf values and calculate:
ΔH = [(25 × ΔHf(CO2)) + (26 × ΔHf(H2O))] - [(1 × ΔHf(C25H52)) + (38 × ΔHf(O2))]
ΔH = [(25 × -393.5) + (26 × -285.8)] - [(1 × 0) + (38 × 0)]
ΔH = (-9837.5 + (-7434.8)) - 0
ΔH = -17272.3 kJ
Therefore, the change in enthalpy for the reaction is approximately -17272.3 kJ.