Standard enthaply of reaction

[zeshkani]

this is my question , i just don't know how to get the standard enthaply of reaction
of this problem iam sure its easy, but i just don't get it, any help is welcome

14. Given the reactions (1) and (2) below, determine (a) ΔrH and ΔrU for reaction (3), (b) ΔrH for both HBr(g) and H2O(g) all at 298 K.
(1) H2(g) + Br2(l) → 2 HBr(g) ΔrH = - 72.80 kJ/mol
(2) 2 H2(g) + O2(g) → 2 H2O(g) ΔrH = –483.64 kJ/mol
(3) 4 HBr(g) + O2(g) → 2 Br2(l) + 2 H2O(g)

 

[Nerro]

Look at reaction (3) as two separate phases, first reaction (1) happens in reverse and then reaction (2) happens. This isn't what happens but all the energies involved here are state functions which means that it doesn't matter how you got to that state, the energy is always the same. This way complex problems can be simplified.

PS. This also works for example when you want to calculate the difference in volume of a gas when you both heat it and change the pressure. First you calculate the change in volume due to the pressure change and than with the new volume you do the same with the new temperature (or vise versa, it doesn't matter). It's neat, give it a go ;)

 

[Blueshift5]

First, we need to understand what the standard enthaply of reaction (ΔrH) is. It is the change in enthalpy that occurs during a chemical reaction under standard conditions (1 atm pressure, 298 K temperature, and 1 mol concentration). It is a measure of the heat absorbed or released during a reaction.

To calculate ΔrH for reaction (3), we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction. In this case, we can break down reaction (3) into two steps: (1) the formation of HBr from H2 and Br2, and (2) the combustion of HBr to form H2O and Br2.

Step 1: H2(g) + Br2(l) → 2 HBr(g) ΔrH = - 72.80 kJ/mol
Step 2: 2 HBr(g) + O2(g) → 2 Br2(l) + 2 H2O(g) ΔrH = -483.64 kJ/mol

Adding these two steps together, we get:
4 HBr(g) + O2(g) → 2 Br2(l) + 2 H2O(g) ΔrH = -556.44 kJ/mol

Therefore, the standard enthalpy of reaction for (3) is -556.44 kJ/mol.

To calculate ΔrU, we need to use the relationship ΔrU = ΔrH – ΔnRT, where Δn is the change in moles of gas and R is the gas constant (8.314 J/mol•K). In this case, Δn = 2 – 4 = -2, and at 298 K, ΔrU = -556.44 kJ/mol - (-2 mol)(8.314 J/mol•K)(298 K) = -554.78 kJ/mol.

For part (b), to find ΔrH for HBr(g) and H2O(g), we can use the same approach. For HBr(g), we can use reaction (1) and for H2O(g), we can use reaction (2). Therefore, ΔrH for HBr(g) is -72.80