Use Hess's law to calculate the enthalpy change for the reaction:
3C(s) + 3H2(g) ¡æ C3H6(g)
Given the following thermochemical equations:
2C3H6(g) + 9O2(g) ¡æ 6CO2(g) + 6H2O(l) ¥ÄH = -4116.0 kJ/mol
C(s) + O2(g) ¡æ CO2(g) ¥ÄH = -393.51 kJ/mol
H2(g) + ¨öO2(g) ¡æ H2O(l) ¥ÄH = -285.83 kJ/mol
please im soo confused!
I can't decipher the symbols, especially the coefficient for oxygen in the fourth equation (counting all equations you posted).
To calculate the enthalpy change for a reaction using Hess's law, you need to manipulate the given thermochemical equations such that when they are added or subtracted, they result in the desired reaction.
Let's break down the desired reaction into the steps needed using the given equations:
Step 1: Combining equations to get the desired reaction
a) Reverse the second equation:
CO2(g) → C(s) + O2(g) ∆H = +393.51 kJ/mol (reversed sign)
b) Multiply the first equation by 2:
4C3H6(g) + 18O2(g) → 12CO2(g) + 12H2O(l) ∆H = -8232.0 kJ/mol (multiplied by 2)
c) Multiply the second equation by 3:
3CO2(g) → 3C(s) + 3O2(g) ∆H = +1180.53 kJ/mol (multiplied by 3)
d) Multiply the third equation by 3:
3H2(g) + 3/2O2(g) → 3H2O(l) ∆H = -857.49 kJ/mol (multiplied by 3)
Step 2: Combine the manipulated equations to obtain the desired reaction:
Add the manipulated equations together:
4C3H6(g) + 18O2(g) + 3CO2(g) + 3H2(g) + 3/2O2(g) → 12CO2(g) + 12H2O(l) + 3C(s) + 3O2(g) + 3H2O(l)
Simplifying the equation:
4C3H6(g) + 15O2(g) → 12CO2(g) + 15H2O(l) + 3C(s)
Step 3: Calculate the overall ∆H
Add up the ∆H values of the manipulated equations:
∆H = (-8232.0 kJ/mol) + (+1180.53 kJ/mol) + (-857.49 kJ/mol)
= -8232.0 kJ/mol + 1180.53 kJ/mol - 857.49 kJ/mol
= -7437.96 kJ/mol
Therefore, the enthalpy change (∆H) for the given reaction 3C(s) + 3H2(g) → C3H6(g) is -7437.96 kJ/mol.