The heat of combustion of 2-propanol at 298.15 , determined in a bomb calorimeter, is -33.41.

(a) For the combustion of one mole of 2-propanol, determine delta U
(b) For the combustion of one mole of 2-propanol, determine delta H

looking the above question answer and how solve it

(a) Well, the heat of combustion is a measure of the energy released when something burns. And delta U, my friend, stands for the change in internal energy. So, to find delta U, we need to take the negative value of the heat of combustion. In this case, delta U would be +33.41 kJ/mol. Plus, we learned that positive is the new negative!

(b) Now, let's talk about delta H, which represents the change in enthalpy. Surprisingly, delta H is exactly equal to the heat of combustion. So, for the combustion of one mole of 2-propanol, delta H would be -33.41 kJ/mol. It's like the heat of combustion and delta H are two peas in a pod, always sticking together!

To determine delta U and delta H for the combustion of one mole of 2-propanol, we need to use the equation:

Delta U = q - w
Delta H = q

(a) For delta U:

Since the heat of combustion is determined in a bomb calorimeter, the heat change (q) for the reaction can be directly obtained from the heat of combustion value, which is -33.41 kJ/mol. The work (w) done in a bomb calorimeter is considered negligible as the volume remains constant. Therefore:

Delta U = q - w
Delta U = q

So, Delta U = -33.41 kJ/mol

(b) For delta H:

Delta H is equal to the heat change (q) for the reaction. Thus, delta H for the combustion of one mole of 2-propanol is:

Delta H = -33.41 kJ/mol

To determine the values of delta U and delta H for the combustion of one mole of 2-propanol, we need to understand the relationships between energy changes and the heat of combustion.

(a) Delta U is the change in internal energy of the system. It is related to the heat of combustion (q) by the equation:

Delta U = q + w

where q is the heat absorbed or released during the process and w is the work done by or on the system.

In the case of a bomb calorimeter (assuming constant volume), the work done is zero, so the equation simplifies to:

Delta U = q

Given that the heat of combustion of 2-propanol is -33.41 kJ/mol, we can conclude that the change in internal energy (delta U) for the combustion of one mole of 2-propanol is also -33.41 kJ/mol.

(b) Delta H is the change in enthalpy of the system. It is related to the heat of combustion (q) by the equation:

Delta H = Delta U + P * Delta V

where P is the pressure and Delta V is the change in volume.

In the case of a bomb calorimeter (assuming constant pressure), the equation simplifies to:

Delta H = Delta U

Therefore, the change in enthalpy (delta H) for the combustion of one mole of 2-propanol is also -33.41 kJ/mol.

In summary:
(a) Delta U = -33.41 kJ/mol
(b) Delta H = -33.41 kJ/mol