AT 25 DEGREES CELSIUS THE STANDARD ENTHALPY OF COMBUSTION OF GASEOUS PROPANE IS -2219.0 KJ PER MOLE OF PROPANE AND THER STANDARD ENTHALPY OF GASEOUS PORPYLENE IS -2058.3 KJ PER MOLE OF P[ROPYLENE. wHAT IS THE STANDARD ENTHALPY CHANGE FOR THE FOLLOWING REACTION C3H6 + H2 ---C3H8
The equation is
DHrxn = (n*DHfproducts) - (n*DHfreactants)
Balanced the equation (I think it is balanced as is). DH stands for delta H and DHf is delta Hformation. You can find DHf information for 25C probably in the back of your text. Note: This method does NOT use the information in the problem; i.e., it does not use the heat of combustion.
To find the standard enthalpy change for the reaction C3H6 + H2 ⟶ C3H8, we need to use Hess's law, which states that the total enthalpy change of a reaction is independent of the path taken as long as the initial and final conditions are the same.
We can begin by writing the given equations:
1. C3H6 ⟶ C3H8 (Standard Enthalpy Change: ?)
2. H2 ⟶ 0.5H2 (Standard Enthalpy Change: Half of the standard enthalpy of combustion of gaseous hydrogen)
Now, let's manipulate the given equations so that they add up to the desired reaction:
1. C3H6 ⟶ C3H8
Since we want C3H8 on the right side and C3H6 on the left side, we need to reverse equation 1. This will also change the sign of the enthalpy change.
C3H8 ⟶ C3H6 (Standard Enthalpy Change: +∆H1)
2. H2 ⟶ 0.5H2
We need 2 moles of H2 on the left side to balance the equation, so we multiply equation 2 by 2 and halve the enthalpy change since we are using half the amount of H2.
2H2 ⟶ H2 (Standard Enthalpy Change: -0.5 * Entropy of combustion of gaseous hydrogen)
Now, add the equations together to obtain the overall reaction:
C3H6 + H2 ⟶ C3H8
C3H8 ⟶ C3H6 (Standard Enthalpy Change: +∆H1)
2H2 ⟶ H2 (Standard Enthalpy Change: -0.5 * Entropy of combustion of gaseous hydrogen)
C3H6 + 3H2 ⟶ C3H8 (Overall Reaction)
Since we have canceled out the H2 terms, we only have C3H6 and C3H8 remaining in the overall reaction equation. The total enthalpy change will be the sum of the individual enthalpy changes (∆H1 + -0.5 * Entropy of combustion of gaseous hydrogen).
So, the Standard Enthalpy Change (∆H) for the reaction C3H6 + H2 ⟶ C3H8 is +∆H1 + -0.5 * Entropy of combustion of gaseous hydrogen.
To calculate the standard enthalpy change (ΔH°) for the given reaction, we need to consider the standard enthalpies of combustion for gaseous propane (C3H8) and gaseous propylene (C3H6).
The reaction you provided is:
C3H6 + H2 → C3H8
First, let's write down the equations for the complete combustion of propane (C3H8) and propylene (C3H6).
Complete combustion of propane:
C3H8 + 5O2 → 3CO2 + 4H2O
Complete combustion of propylene:
C3H6 + 4.5O2 → 3CO2 + 3H2O
We can see that the reaction we want to calculate is going from propylene (C3H6) to propane (C3H8), which means we need to reverse and adjust the combustion equations to match the given reaction.
Reversed combustion of propane:
3CO2 + 4H2O → C3H8 + 5O2
Reversed combustion of propylene:
3CO2 + 3H2O → C3H6 + 4.5O2
Now, we can calculate the enthalpy changes for the reversed combustion equations using the given standard enthalpy values.
Enthalpy change for reversed combustion of propane:
ΔH1° = -(-2219.0 kJ/mol) = 2219.0 kJ/mol
Enthalpy change for reversed combustion of propylene:
ΔH2° = -(-2058.3 kJ/mol) = 2058.3 kJ/mol
Since we reversed the equations, the signs of the enthalpy changes also change.
Now, we can add the enthalpy changes for the reversed reactions to get the overall enthalpy change for the given reaction:
ΔH° = ΔH1° - ΔH2°
= 2219.0 kJ/mol - 2058.3 kJ/mol
= 160.7 kJ/mol
Therefore, the standard enthalpy change for the reaction C3H6 + H2 → C3H8 is 160.7 kJ/mol.